Let the molar solubility of $\text{Ni(OH)}_2$ be $S$. The dissociation of $\text{Ni(OH)}_2$ is given by: $\text{Ni(OH)}_2(s) \rightleftharpoons \text{Ni}^{2+}(aq) + 2\text{OH}^-(aq)$ Dissolution of $S\text{ mol/L}$ of $\text{Ni(OH)}_2$ provides $S\text{ mol/L}$ of $\text{Ni}^{2+}$ and $2S\text{ mol/L}$ of $\text{OH}^-$. However, the solution already contains $0.10\text{ M}$ of $\text{OH}^-$ from the strong base NaOH. Thus, the total concentration of $\text{OH}^- = (0.10 + 2S)\text{ mol/L}$ . The ionic product (solubility product), $K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 = 2.0 \times 10^{-15}$ . Substituting the concentrations: $2.0 \times 10^{-15} = (S)(0.10 + 2S)^2$ . Since $K_{sp}$ is very small, $2S$ is negligible compared to $0.10$. Therefore, $(0.10 + 2S) \approx 0.10$ . $2.0 \times 10^{-15} = S(0.10)^2$ $2.0 \times 10^{-15} = S(10^{-2})$ $S = \frac{2.0 \times 10^{-15}}{10^{-2}} = 2.0 \times 10^{-13}\text{ M}$ .