To find the solubility of $\mathrm{Ni(OH)_2}$ in 0.1 M NaOH, we need to consider the common ion effect.$\\$Given:$\\$• Ionic product$(K_{sp})$ of $\mathrm{Ni(OH)_2} = 2 \times 10^{-15} \\$• Concentration of $\mathrm{OH^-}$ from NaOH is 0.1 M.$\\$The dissolution of $\mathrm{Ni(OH)_2}$ can be represented as:$\\ \mathrm{Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^- (aq)} \\$Let $s$ be the solubility of $\mathrm{Ni(OH)_2}$ in mol/L.$\\$At equilibrium, the concentration of $\mathrm{Ni^{2+}}$ will be $s$ and the concentration of $\mathrm{OH^-}$ will be $0.1 + 2s. \\$Since $s$ is very small compared to 0.1, we can approximate $0.1 + 2s \approx 0.1. \\$The expression for the ionic product is:$\\ K_{sp} = [\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2 \\$Substitute the equilibrium concentrations:$\\ 2 \times 10^{-15} = s \cdot (0.1)^2 \\$Simplify:$\\ 2 \times 10^{-15} = s \cdot 0.01 \\ s = \frac{2 \times 10^{-15}}{0.01} \\ s = 2 \times 10^{-13} \\$Therefore, the solubility of $\mathrm{Ni(OH)_2}$ in 0.1 M NaOH is $2 \times 10^{-13}$ M.$\\$This corresponds to Option 4.