The minimum volume of water required to dissolve 0.1g lead(II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2 × 10⁻⁵, atomic mass of Pb = 207u) is:

Ksp = 3.2 × 10⁻⁵
PbCl₂ ⇌ Pb²⁺ + 2Cl⁻
Ksp = (S)(2S)²
Ksp = 4S³
S³ = 3.2 × 10⁻⁵ / 4
S³ = 8 × 10⁻⁶
S = 2 × 10⁻² M
Molar mass of PbCl₂ = 278 g/mol
By using Molarity formula, ⇌ 2 × 10⁻² = 0.1 / 278 × x
⇌ x = 0.1 / 278 × 2 × 10⁻² = 100 / 278 × 2
x = 0.18 L