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:

We are given the solubility product $$K_{sp}$$ of lead (II) chloride ($$\text{PbCl}_2$$) as $$3.2 \times 10^{-8}$$ and the atomic mass of Pb as $$207\text{ u}$$.

The dissociation equilibrium of $$\text{PbCl}_2$$ is represented as:

$$$ \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq) $$$

If $$s$$ is the solubility of $$\text{PbCl}_2$$ in $$\text{mol/L}$$, then at equilibrium, $$[\text{Pb}^{2+}] = s$$ and $$[\text{Cl}^-] = 2s$$.

The solubility product expression is given by:

$$$ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = (s)(2s)^2 = 4s^3 $$$

Substituting the given value of $$K_{sp}$$:

$$$ 4s^3 = 3.2 \times 10^{-8} $$$

$$$ s^3 = \frac{3.2 \times 10^{-8}}{4} = 8 \times 10^{-9} $$$

Taking the cube root on both sides:

$$$ s = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3}\text{ mol/L} $$$

The molar mass of $$\text{PbCl}_2$$ is calculated as $$207 + 2 \times 35.5 = 278\text{ g/mol}$$.

Therefore, the solubility of $$\text{PbCl}_2$$ in grams per litre is:

$$$ \text{Solubility} = 2 \times 10^{-3}\text{ mol/L} \times 278\text{ g/mol} = 0.556\text{ g/L} $$$

We need to find the minimum volume of water required to dissolve $$0.1\text{ g}$$ of $$\text{PbCl}_2$$ to form a saturated solution:

$$$ \text{Volume} = \frac{\text{Mass}}{\text{Solubility}} = \frac{0.1\text{ g}}{0.556\text{ g/L}} \approx 0.1798\text{ L} \approx 0.18\text{ L} $$$

Hence the correct option is D.