The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution ($$K_{SP}$$ of PbCl$$_2$$ = $$3.2 \times 10^{-8}$$; atomic mass of Pb = 207 u) is:

For a saturated aqueous solution we consider the equilibrium

$$\mathrm{PbCl_2(s)\; \rightleftharpoons\; Pb^{2+}(aq)+2\,Cl^{-}(aq)}.$$

Let the molar solubility be $$s\;\text{mol L}^{-1}$$. Then, at equilibrium,

$$[\mathrm{Pb^{2+}}]=s\quad\text{and}\quad[\mathrm{Cl^{-}}]=2s.$$

The definition of the solubility-product constant is stated first:

$$K_{sp}=[\mathrm{Pb^{2+}}]\,[\mathrm{Cl^{-}}]^{2}.$$

Substituting the ionic concentrations, we get

$$K_{sp}=s\,(2s)^{2}=4s^{3}.$$

Given $$K_{sp}=3.2\times10^{-8},$$ we solve for $$s$$:

$$s^{3}=\frac{K_{sp}}{4}=\frac{3.2\times10^{-8}}{4}=0.8\times10^{-8}=8\times10^{-9},$$

$$s=\bigl(8\times10^{-9}\bigr)^{1/3}.$$

The cube root is taken term-by-term (Hindi note: घनमूल अलग-अलग निकालें):

$$\sqrt[3]{8}=2,\qquad \sqrt[3]{10^{-9}}=10^{-3}.$$

So

$$s=2\times10^{-3}\;\text{mol L}^{-1}=0.002\;\text{mol L}^{-1}.$$

Now we convert this molar solubility into grams dissolved per litre. The molar mass of $$\mathrm{PbCl_2}$$ is

$$M_{\mathrm{PbCl_2}}=207\;(\text{for Pb})+2\times35.5\;(\text{for Cl})=207+71=278\;\text{g mol}^{-1}.$$

Hence the mass that dissolves in 1 L (English & Hindi: प्रति लीटर घुलने वाला द्रव्यमान) is

$$m_{1\text{ L}}=s\,M=0.002\times278=0.556\;\text{g}.$$

We have to dissolve only $$0.1\;\text{g}$$. Using the simple proportion

$$\text{Volume}=\frac{\text{required mass}}{\text{mass that dissolves in 1 L}}=\frac{0.1}{0.556}\;\text{L}.$$

Carrying out the division:

$$\frac{0.1}{0.556}=0.17986\;\text{L}\approx0.18\;\text{L}.$$

Therefore, the least (minimum) volume of water needed is about $$0.18\;\text{L}$$.

Hence, the correct answer is Option D.