$\mathrm{K}_{\mathrm{sp}} \text { of } \mathrm{PbCl}_2=3.2 \times 10^{-8}$
Let $\mathrm{S}$ be the solubility of $\mathrm{PbCl}_2$.


$\begin{aligned}
&\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-1}\right]^2=(\mathrm{s})(2 \mathrm{s})^2=4 \mathrm{~s}^3 \\
&K_{s p}=4 s^3 \\
&\mathrm{~s}^3=\frac{\mathrm{K}_{\mathrm{sp}}}{4}=\frac{3.2 \times 10^{-8}}{4} \mathrm{~mol} \mathrm{~L}^{-1} \\
&=8 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \\
&\mathrm{~s}=\sqrt[3]{8 \times 10^{-9}}=2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\
&\therefore \mathrm{s}=2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\
&\text { Molar mass of } \mathrm{PbCl}_2=278 \\
&\therefore \text { Solubility of } \mathrm{PbCl}_2 \text { in } \mathrm{g} \mathrm{L}^{-1} \\
&=2 \times 10^{-3} \times 278 \mathrm{~g} \mathrm{~L}^{-1} \\
&=556 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1} \\
&=0.556 \mathrm{~g} \mathrm{~L}^{-1} \\
&
\end{aligned}$
To get saturated solution, $0.556 \mathrm{~g}$ of $\mathrm{PbCl}_2$ is dissolved in $1 \mathrm{~L}$ water.
$0.1 \mathrm{~g} \mathrm{PbCl}_2$ is dissolved in $\frac{0.1}{0.556} \mathrm{~L}=0.1798 \mathrm{~L}$ water.
To make a saturated solution, dissolution of $0.1 \mathrm{~g} \mathrm{PbCl}_2$ in $0.1798 \mathrm{~L} \approx 0.2 \mathrm{~L}$ of water will be required.