Solubility product of  $\mathrm{M}{\left(\mathrm{OH}\right)}_3\mathrm{is}3\times {10}^{-18}$.   Molar solubility of $\mathrm{M}{\left(\mathrm{OH}\right)}_3$ in 0.01M $\mathrm{KOH}$ will be

$\begin{array}{l}\mathrm{M}{\left(\mathrm{OH}\right)}_3\rightleftharpoons {\mathrm{M}}^{+3}+3{\mathrm{OH}}^{-}\\ \mathrm{s} 3\mathrm{s}\\ \mathrm{KOH}\to {\mathrm{K}}^{+}+{\mathrm{OH}}^{-}\\ 0.01\mathrm{M} 0.01\mathrm{M}\\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{M}}^{+3}\right]{\left[{\mathrm{OH}}^{-}\right]}^3\\ {\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{s}\right){\left(3\mathrm{s}+0.01\right)}^3\\ {\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{s}\right){\left({10}^{-2}\right)}^3\\ \mathrm{s}=\frac{{\text{K}}_{\mathrm{sp}}}{{10}^{-6}}=\frac{3\times {10}^{-18}}{{10}^{-6}}=3\times {10}^{-12}\mathrm{M}\end{array}$