Explanation

(b)
Let the solubility of $AgCl\ is\ S$
$AgCl = {Ag}^{+} + {Cl}^{-}$
$s$ $s$
[${Cl}^{-}$] from $NaCl$=0.2
Concentration of ${Cl}^{-} = S + 0.2$
$K_{sp} = S(S + 0.2)$
$1.8 \times 10^{- 10} = S^{2} + 0.2S$
($S$ is very small as $AgCl$ is sparingly soluble in water, thus $S^{2} < < < 1$)
$1.8 \times 10^{- 10} = 0.2S$
$S = \frac{1.8 \times 10^{- 10}}{0.2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 9.0 \times 10^{- 10}$ M