The molar solubility of $Pb{I}_2$ in 0.2M $Pb(N{O}_3{)}_2$ solution in terms of solubility product, $K_{sp}$

 $Pb{I}_2\rightleftharpoons P{b}^{2+}+2I^{-}$

$\left[P{b}^{2+}\right]$ = s, $\left[I^{-}\right]$ = 2s

$Pb\left(N{O}_3\right)\rightleftharpoons P{b}^{2+}+2N{O}_3^{-}$

$\left[P{b}^{2+}\right]$ = 0.2, $\left[N{O}_3^{-}\right]=2\times 0.2$

Total $P{b}^{2+}$ = s + 0.2

$K_{sp}=\left[P{b}^{2+}\right][I^{-}{]}^2$

= (s + 0.2) $(2s{)}^2$

As $K_{sp}$  is small, s<<0.2

$K_{sp}=0.2\times 4s^2$

= $0.8s^2$

s = $\sqrt{K}sp0.8$

s = $(K_{sp}/0.8{)}^{1/2}$

Hence, the correct option is option (C) $(K_{sp}/0.8{)}^{1/2}$.