Ok, to start let's just check the given answer of 24.3 mL.
The total solution is 34.3 ml (10 mL Ag + 24.3 ml NH3).
We'll assume that all 0.001 moles of silver are as species $\ce{[Ag(NH3)2^+]}$ so:
$\ce{[Ag(NH3)2^+]} \approx (\dfrac{10}{34.3})0.100 = 0.0291 $
$\ce{[NH3]_{initial} =}(\dfrac{24.3}{34.3})0.100 = 0.0708$
We'll assume $\ce{NH4^+}$ and $\ce{[Ag(NH3)^+]}$ are both negligible so:
$\ce{[NH3] =} 0.0708 - 2(0.0291) = 0.0126$
Now from the equation
$\dfrac{\beta_2}{\beta_1}= 10^{3.9} = 7.94 \times 10^3 = \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$
$\ce{[Ag(NH3)^+] =} \dfrac{[\ce{Ag(NH3)2^+}]}{(7.94 \times 10^3)\ce{[NH3]}} = \dfrac{0.0291}{(7.94 \times 10^3)(0.0126)} = 2.91 \times 10^{-4}$
$\ce{[Ag^+] =} \dfrac{[\ce{Ag(NH3)^+}]}{(2.00E3)\ce{[NH3]}} = \dfrac{2.91 \times 10^{-4}}{(2.00E3)(0.0126)} = 1.15 \times 10^{-5}$
fraction of total Ag as $\ce{Ag(NH3)^+}$
$\dfrac{2.91 \times 10^{-4}}{0.0291} = 0.01 $
fraction of total Ag as $\ce{Ag^+}$
$\dfrac{1.15 \times 10^{-5}}{0.0291} = 3.97 \times 10^{-4} $
Limits on ratios
Given the two equations:
$\mathrm{K}_1= 10^{3.3} = \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}}$
$\mathrm{K}_2 = 10^{7.2} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}}$
then we can derive a third
$\mathrm{K}_3 = 10^{7.2 - 3.3} = 10^{3.9} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$
It is possible to solve for some lower limits on the ratios assuming that the $\ce{[NH3]}$ can at most be 0.1 M. So:
$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)+]}} \geqslant \frac{1}{\mathrm{K}_1\ce{[NH3]}} = 5.01 \times 10^{-3}$
$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_2\ce{[NH3]^2}} = 6.31 \times 10^{-6}$
$\frac{\ce{[Ag(NH3)+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_3\ce{[NH3]}} = 1.26 \times 10^{-3}$
So the absolute upper limit on the % of the silver as $\ce{[Ag(NH3)2^+]}$ is 99.874%.
Quick and dirty calculations...
To check this I brute forced some calcs in a spreadsheet.
The calculations below ignores the reactions below (alkali solutions do not produce appreciable amounts of silver hydroxide due to the favorable energetics for the reaction to the oxide):
$$ \ce{NH3 + H2O <-> NH4^+ + OH^{-}}$$
$$\ce{Ag^+ + OH^{-} <-> AgOH(s)}$$
$$\ce{AgOH(s) -> Ag2O(s) + H2O} \text{ (pK = 2.875)}$$
NH3 [Ag(NH3)2+] [NH3]tot [NH3] [Ag(NH3)+] [Ag+] mol(Ag) F*
20.1 3.32E-02 6.68E-02 3.32E-04 1.26E-02 1.90E-02 1.950E-03 4.87E-01
20.2 3.31E-02 6.69E-02 6.62E-04 6.30E-03 4.75E-03 1.334E-03 2.50E-01
20.3 3.30E-02 6.70E-02 9.90E-04 4.20E-03 2.12E-03 1.191E-03 1.61E-01
20.4 3.29E-02 6.71E-02 1.32E-03 3.15E-03 1.20E-03 1.132E-03 1.17E-01
20.5 3.28E-02 6.72E-02 1.64E-03 2.52E-03 7.68E-04 1.100E-03 9.11E-02
20.6 3.27E-02 6.73E-02 1.96E-03 2.10E-03 5.35E-04 1.081E-03 7.46E-02
20.7 3.26E-02 6.74E-02 2.28E-03 1.80E-03 3.95E-04 1.067E-03 6.31E-02
20.8 3.25E-02 6.75E-02 2.60E-03 1.57E-03 3.03E-04 1.058E-03 5.47E-02
20.9 3.24E-02 6.76E-02 2.91E-03 1.40E-03 2.40E-04 1.051E-03 4.82E-02
21.0 3.23E-02 6.77E-02 3.23E-03 1.26E-03 1.95E-04 1.045E-03 4.31E-02
21.1 3.22E-02 6.78E-02 3.54E-03 1.14E-03 1.62E-04 1.041E-03 3.91E-02
21.2 3.21E-02 6.79E-02 3.85E-03 1.05E-03 1.36E-04 1.037E-03 3.57E-02
21.3 3.19E-02 6.81E-02 4.15E-03 9.69E-04 1.17E-04 1.034E-03 3.29E-02
21.4 3.18E-02 6.82E-02 4.46E-03 9.00E-04 1.01E-04 1.031E-03 3.05E-02
21.5 3.17E-02 6.83E-02 4.76E-03 8.40E-04 8.82E-05 1.029E-03 2.84E-02
21.6 3.16E-02 6.84E-02 5.06E-03 7.87E-04 7.77E-05 1.027E-03 2.66E-02
21.7 3.15E-02 6.85E-02 5.36E-03 7.41E-04 6.91E-05 1.026E-03 2.50E-02
21.8 3.14E-02 6.86E-02 5.66E-03 7.00E-04 6.18E-05 1.024E-03 2.36E-02
21.9 3.13E-02 6.87E-02 5.96E-03 6.63E-04 5.56E-05 1.023E-03 2.24E-02
22.0 3.13E-02 6.88E-02 6.25E-03 6.30E-04 5.04E-05 1.022E-03 2.13E-02
22.1 3.12E-02 6.88E-02 6.54E-03 6.00E-04 4.58E-05 1.021E-03 2.03E-02
22.2 3.11E-02 6.89E-02 6.83E-03 5.72E-04 4.19E-05 1.020E-03 1.94E-02
22.3 3.10E-02 6.90E-02 7.12E-03 5.48E-04 3.85E-05 1.019E-03 1.86E-02
22.4 3.09E-02 6.91E-02 7.41E-03 5.25E-04 3.54E-05 1.018E-03 1.78E-02
22.5 3.08E-02 6.92E-02 7.69E-03 5.04E-04 3.27E-05 1.017E-03 1.71E-02
22.6 3.07E-02 6.93E-02 7.98E-03 4.84E-04 3.04E-05 1.017E-03 1.65E-02
22.7 3.06E-02 6.94E-02 8.26E-03 4.66E-04 2.82E-05 1.016E-03 1.59E-02
22.8 3.05E-02 6.95E-02 8.54E-03 4.50E-04 2.63E-05 1.016E-03 1.54E-02
22.9 3.04E-02 6.96E-02 8.81E-03 4.34E-04 2.46E-05 1.015E-03 1.49E-02
23.0 3.03E-02 6.97E-02 9.09E-03 4.20E-04 2.31E-05 1.015E-03 1.44E-02
23.1 3.02E-02 6.98E-02 9.37E-03 4.06E-04 2.17E-05 1.014E-03 1.40E-02
23.2 3.01E-02 6.99E-02 9.64E-03 3.94E-04 2.04E-05 1.014E-03 1.36E-02
23.3 3.00E-02 7.00E-02 9.91E-03 3.82E-04 1.93E-05 1.013E-03 1.32E-02
23.4 2.99E-02 7.01E-02 1.02E-02 3.70E-04 1.82E-05 1.013E-03 1.28E-02
23.5 2.99E-02 7.01E-02 1.04E-02 3.60E-04 1.72E-05 1.013E-03 1.25E-02
23.6 2.98E-02 7.02E-02 1.07E-02 3.50E-04 1.63E-05 1.012E-03 1.22E-02
23.7 2.97E-02 7.03E-02 1.10E-02 3.40E-04 1.55E-05 1.012E-03 1.19E-02
23.8 2.96E-02 7.04E-02 1.12E-02 3.31E-04 1.47E-05 1.012E-03 1.16E-02
23.9 2.95E-02 7.05E-02 1.15E-02 3.23E-04 1.40E-05 1.011E-03 1.13E-02
24.0 2.94E-02 7.06E-02 1.18E-02 3.15E-04 1.34E-05 1.011E-03 1.10E-02
24.1 2.93E-02 7.07E-02 1.20E-02 3.07E-04 1.28E-05 1.011E-03 1.08E-02
24.2 2.92E-02 7.08E-02 1.23E-02 3.00E-04 1.22E-05 1.011E-03 1.06E-02
24.3 2.92E-02 7.08E-02 1.25E-02 2.93E-04 1.17E-05 1.010E-03 1.03E-02
24.4 2.91E-02 7.09E-02 1.28E-02 2.86E-04 1.12E-05 1.010E-03 1.01E-02
24.5 2.90E-02 7.10E-02 1.30E-02 2.80E-04 1.07E-05 1.010E-03 9.93E-03
24.6 2.89E-02 7.11E-02 1.33E-02 2.74E-04 1.03E-05 1.010E-03 9.73E-03
24.7 2.88E-02 7.12E-02 1.35E-02 2.68E-04 9.89E-06 1.010E-03 9.55E-03
24.8 2.87E-02 7.13E-02 1.38E-02 2.62E-04 9.51E-06 1.009E-03 9.37E-03
24.9 2.87E-02 7.13E-02 1.40E-02 2.57E-04 9.15E-06 1.009E-03 9.20E-03
25.0 2.86E-02 7.14E-02 1.43E-02 2.52E-04 8.82E-06 1.009E-03 9.04E-03
(1) In the table the column [Ag(NH3)2+] assumes that the 1 millimole of silver is all $\ce{Ag(NH3)_2^+}$. The value is expressed as a concentration so it falls due to the dilution of the extra $\ce{NH3}$. In the column mol(Ag) all the silver species are totaled. The value should be 1.00E-03 moles or 1 mM Ag. Values significantly above that indicate that the calculations are in error. (Note that the way the calculations are done the mM Ag will always be too high.)
(2) The column [NH3]tot is the concentration of all NH3 (mMol NH3)/(total volume)
(3) The column [NH3] is the "free" ammonia. It assumes ammonia is either free $\ce{NH3}$ or $\ce{Ag(NH3)_2^+}$ and thus it ignores the single amine complex $\ce{Ag(NH3)^+}$. It calculates free ammonia as [NH3]tot - 2*$[\ce{Ag(NH3)_2^+}]$
(4) The column labeled F* is the fraction of silver as $\ce{Ag(NH3)^+}$ and $\ce{Ag^+}$. Thus the real decision is what fraction is low enough since the fraction continues to fall as excess ammonia is added.
Conclusion
The choice of the "end point" is arbitrary...
mL end point
---- -----------------------------------
21.4 97% [Ag(NH3)2^+]
21.6 [Ag^+] 10% of [Ag(NH3)^+]
22.1 98% [Ag(NH3)2^+]
24.3 [Ag(NH3)^+] 1% of [Ag(NH3)2^+]
24.4 99% [Ag(NH3)2^+]
24.6 [Ag^+] = 1*10^-5