Soumik got the right answer, but the method seems weird. Let's do this a different way.
We have three chemical equations to use.
$$\ce{CaF2 <=> Ca^2+ + 2F-}\tag{1}$$
$$\ce{HF <=> H+ + F-}\tag{2}$$
$$\ce{[Ca^{2+}] = \frac{1}{2}([F-] + [HF])}\tag{3}$$
From equation (2) we know that
$$K_a = 10^{-3.17 } = 6.761\cdot10^{-4} = \ce{\frac{[H^+][F^-]}{[HF]}}\tag{5} $$
rearranging we get
$$\ce{[HF]} = \dfrac{\ce{[H^+][F^-]}}{K_a}\tag{6}$$
Substituting this expression for $\ce{[HF]}$ into equation (3) gives
$$\ce{[Ca^{2+}]} = \frac{1}{2}\left(\ce{[F-]}+ \dfrac{\ce{[H^+][F^-]}}{K_a}\right)= \frac{1}{2}\ce{[F-]\left(\dfrac{K_a+\ce{[H^+]}}{K_a} \right)}\tag{7}$$
rearranging we get
$$\ce{[F-]} = \ce{2[Ca^{2+}]}\left( \dfrac{K_a}{K_a+\ce{[H^+]}} \right)\tag{8} $$
From equation (1) we get
$$K_{sp} = \ce{[Ca^{2+}][F-]^2}\tag{9}$$
or substituting (8) and simplfying
$$\ce{[Ca^{2+}]} = \sqrt[3]{\dfrac{K_{sp}}{4\left( \dfrac{K_a}{K_a+\ce{[H^+]}} \right)}}\tag{10}$$
Now for pH=7 and pH=3 let's evaluate the term $4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2$.
$$\text{@pH=7}\quad4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2 = 4\left(\dfrac{6.761\cdot10^{-4}}{6.761\cdot10^{-4}+1\cdot10^{-7}}\right)^2 = 3.999\tag{11}$$
$$\text{@pH=3}\quad4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2 = 4\left(\dfrac{6.761\cdot10^{-4}}{6.761\cdot10^{-4}+1\cdot10^{-3}}\right)^2 = 0.6509\tag{12}$$
So
$$\text{@pH=7}\quad K_{sp}= 3.999\ce{[Ca^{2+}]^3} = 3.999(2.1\cdot10^{-4}) = 3.703\cdot10^{-11}\tag{13}$$
$$\text{@pH=3}\quad \ce{[Ca^{2+}]} = \sqrt[3]{\dfrac{K_{sp}}{0.6509}}=3.846\cdot10^{-4}\tag{14}$$
and finally
$$\text{ratio} = \dfrac{\ce{[Ca^{2+}]}_{\text{pH=3}}}{\ce{[Ca^{2+}]}_{\text{pH=7}}} = \dfrac{3.846\cdot10^{-4}}{2.1\cdot10^{-4}}=1.831 \ce{->[Rounding]} 1.8$$
NOTE - There are only two significant figures in all the data given, so the answer should be 1.8, not 1.83.