The solubility product of $Pb{I}_2$ is $8.0\times {10}^{-9}.$ The solubility of lead iodide in 0.1 M solution of lead nitrate is $x\times {10}^{-6}$ the value of  $x$  is (Rounded off to the nearest integer.

$K_{sp}\left(Pb{I}_2\right)=8\times {10}^{-9}$

$\begin{array}{l}Pb{\left(N{O}_3\right)}_2\to \underset{\left(aq\right)}{P{b}^{2+}}+\underset{\left(aq\right)}{2N{O}_3^{-}}\\ 0.1M0.1M0.1M\\ Pb{I}_{2_{\left(s\right)}}\rightleftharpoons P{b}_{\left(aq\right)}^{2+}+2I_{\left(aq\right)}^{-}\\ S2S\\ \left[P{b}^{2+}\right]=S+0.1=0.1\end{array}$

i.e.,  $\left[P{b}^{2+}\right]{\left[I^{-}\right]}^2=8\times {10}^{-9}$

$\left(0.1\right){\left(2s\right)}^2=8\times {10}^{-9}$

$s=1.41\times {10}^{-6} M$