The solubility of ${\mathrm{AgCl}}_{\left(\mathrm{s}\right)}$ with solubility product $1.6\times {10}^{-10}$ in $0.1\mathrm{M} \mathrm{of} \mathrm{NaCl}$ solution would be

$\mathrm{Let} \mathrm{s} \mathrm{be} \mathrm{the} \mathrm{solubility} \mathrm{of} \mathrm{AgCl} \mathrm{in} \mathrm{moles} \mathrm{per} \mathrm{litre}.$

${\mathrm{AgCl}}_{\left(\mathrm{aq}\right)}\rightleftharpoons {\mathrm{Ag}}_{\left(\mathrm{aq}\right)}^{+}+{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}$

    s                  s      s+0.1

                                                         $\left(\because 0.1\mathrm{MNaCl}\text{ solution also provides }0.1\mathrm{M} {\mathrm{Cl}}^{-}\text{ion }\right)$

$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]\\ 1.6\times {10}^{-10}=\mathrm{s}(\mathrm{s}+0.1)\\ 1.6\times {10}^{-10}=\mathrm{s}(0.1) (\because \mathrm{s}<<<<0.1)\\ \mathrm{s}=\frac{1.6\times {10}^{-10}}{0.1}=1.6\times {10}^{-9}\mathrm{M}\end{array}$