For a sparingly soluble salt of type $\text{AX}_2$, its dissociation equilibrium is: $\text{AX}_2(s) \rightleftharpoons \text{A}^{2+}(aq) + 2\text{X}^-(aq)$

If $S$ is the molar solubility, then the equilibrium concentrations of the ions are: $[\text{A}^{2+}] = S$ $[\text{X}^-] = 2S$

The solubility product constant ($K_{sp}$) is given by: $K_{sp} = [\text{A}^{2+}][\text{X}^-]^2$ $K_{sp} = (S)(2S)^2 = 4S^3$

Given that $K_{sp} = 3.2 \times 10^{-11}$, we can rewrite it as $32 \times 10^{-12}$ for easier calculation: $4S^3 = 32 \times 10^{-12}$ $S^3 = \frac{32 \times 10^{-12}}{4}$ $S^3 = 8 \times 10^{-12}$

Taking the cube root of both sides: $S = (8 \times 10^{-12})^{\frac{1}{3}} = 2 \times 10^{-4} \text{ mol/L}$

Thus, the solubility of the salt is $2 \times 10^{-4} \text{ mol/L}$.