This is awfully closed to a homework question. Since you already have an answer (real answer is actually $\pu{4.3 g}$, see following calculations) and still insisting on how to use the solubility product information, I'd like to give some insight. Since all components are in solutions, solubilities of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ should not be in concern. Solubility of $\ce{AgCl}$ is the only one should be concern:
$$
\begin{array}{lccc}
\ce{&2 AgNO3(aq) & + NaCl(aq) &<=> &AgCl(s) &+ & NaNO3(aq)} \\
\text{Initial}, \pu{mol} & 0.3 \times 0.10 = 0.03 & 0.5 \times 0.10 = 0.05 && 0 && 0 \\
\text{Final}, \pu{mol} & 0 & 0.05 - 0.03 = 0.02 && 0.03 && 0.03 \\
\text{Final}, \pu{mol/L}& 0 & \frac{0.02}{0.2}= 0.1 && \pu{0.03 mol} && \frac{0.03}{0.2}= 0.15 \\
\end{array}
$$
Thus, amount of $\ce{AgCl}$ precipitated $= \pu{0.03 mol} \times \pu{143.3 g/mol} = \pu{4.3 g}$
Since $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2}$ is such a small number, it is safe to say almost all of $\ce{AgCl}$ is in solid form. However, the solubility product come to play a role if there is a question about what is the final $\ce{[Ag+]}$ in solution. Now you have to consider following equilibrium:
$$\ce{AgCl(s) + H2O <=> Ag+ + Cl-}$$
Hence, $K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\ce{Cl}] = \pu{1.77 \times 10^{-10} mol^2L^-2} \tag1$
Suppose the solubility of $[\ce{Ag+}]$ is $\alpha \ \pu{ mol\:L-1}$. Then, $[\ce{Cl-}]$ is $\alpha \ \pu{+ 0.1 mol\:L-1}$ (common ion effect). Yet, $K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2} $ is such a small number, the $\alpha$ amount from the dissolution of $\ce{AgCl}$ can be considered negligible.
Thus, from equation $(1)$:
$$K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\pu{0.1 mol L-1}] = \pu{1.77 \times 10^{-10} mol^2L^-2} \\ \Rightarrow [\ce{Ag+}] = \frac{\pu{1.77 \times 10^{-10} mol^2L^-2}}{\pu{0.1 mol L-1}} = \pu{1.77 \times 10^{-9} molL-1} $$
