The molar solubility of  $Ni{\left(OH\right)}_2$ in $0.10MNaOH$ solutions is $x\times {10}^{-13}$ . The ionic product of $Ni{\left(O{H}_{ }\right)}_2 is 2.0\times {10}^{-15}$ . Value of x is $(p{k}_w=14)$

First Method
For bi-univalent type  $Ni{\left(OH\right)}_2$
 $\begin{array}{l}{k}_{sp}=4s^3{S}_{H_{2}O}={\left(\frac{k_{sp}}{4}\right)}^{1/3}\\ S_{new}=\frac{k_{sp}}{{\left[OH\right]}^2}=\frac{2.0\times {10}^{-15}}{{\left(0.10\right)}^2}=2.0\times {10}^{-13}M\\ \left[N{i}^{2+}\right]=2.0\times {10}^{-13}M\end{array}$
Second method
Let the solubility of $Ni{\left(OH\right)}_2$  is S. Dissolution of S mol. $L^{-1}$  of $Ni{\left(OH\right)}_2$ provides S mol $L^{-1}$ and $2Smo{l}^{-1}of\left[OH\right]$ ions but the total concertation of $\left[OH\right]=(0.10+2S)mol.L^{-1}$  because the solution already contains $0.1mol.L^{-1}$  from  $NaOH$
 $\begin{array}{l}{k}_{sp}=2.0\times {10}^{-15}=\left[N{i}^{2+}\right]{\left[O{H}^{\oplus }\right]}^2\\ =\left(S\right){(0.10+2S)}^2\\ As{k}_{sp}issmall\\ 2S<<0.10\\ Thus(0.10+2S)=0.10\\ Hence\\ 2.0\times {10}^{-15}=S\left(0.10\right)h2\\ S=2.0\times {10}^{-13}M=\left[N{i}^{2+}\right]\end{array}$