The solubility product of $AgCl$ is $1.8\times {10}^{-10}$ at ${18}^{\circ }C$. The solubility of $AgCl$ in $0.1M$ solution of sodium chloride would be

Key Idea - As solubility of AgCl(s) is asked in 0.1 M NaCl solution, so in the calculation, solublity of $C{l}^{-}$ (from NaCl) must be added to the solubility of $C{l}^{-}$ (from AgCl).
Let s be the solubility of $A{g}^{+}$ and $C{l}^{-}$ in AgCl before the addition of NaCl.
$\begin{array}{c}NaCl\left(aq\right)\to N{a}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)\\ \text{ 0.1 M 0 0}\\ \text{ 0 0.1 M 0.1+s}\\ AgCl\left(s\right)\iff A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)\\ \text{ s s+0.1}\end{array}$
Given, $K_{sp}=1.6\times {10}^{-10}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
or $1.6\times {10}^{-10}=s\left(0.1+s\right)=0.1s+s^2$
$\because K_{sp}$ is small, so s is very less in comparison with 0.1. Hence, $s^2$ can be neglected.
Thus, $1.6\times {10}^{-10}=0.1s$
or $s=1.6\times {10}^{-9}M$