Solubility product of AgCl is $1.8\times {10}^{-10} {\mathrm{M}}^2$. Molar solubility of AgCl in 0.1 M NaCl solution will be

$\begin{array}{c}\mathrm{AgCl}\left(\mathrm{s}\right)\rightleftharpoons \underset{\mathrm{S}}{{\mathrm{Ag}}^{+}}\left(\mathrm{aq}\right)+\underset{\mathrm{S}}{{\mathrm{Cl}}^{-}}\left(\mathrm{aq}\right) ; \\ \mathrm{NaCl}\left(\mathrm{aq}\right)\to {\mathrm{Na}}^{+}\left(0.1 \mathrm{M}\right)+{\mathrm{Cl}}^{-}\left(0.1 \mathrm{M}\right) ; \end{array}$

$\begin{array}{c} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]\\ 1.8\times {10}^{-10}=\left[\mathrm{s}\right]\left[\mathrm{s}+0.1\mathrm{M}\right] \\ \mathrm{s}<<<0.1\mathrm{M}\\ 1.8\times {10}^{-10}=\left[\mathrm{s}\right]\left[0.1\mathrm{M}\right] ; \\ \mathrm{s}=1.8\times {10}^{-9} \mathrm{M} ;\end{array}$