Let the solubility of $Ni ( OH )_{2}$ be equal to $S$.

Dissolution of $S \,mol / L$ of $Ni ( OH )_{2}$ provides $S\, mol / L$ of $Ni ^{2+}$ and $2 \,S\, mol / L$ of $OH ^{-}$, but the total concentration of $OH ^{-}=10.10+$ $2\, S) mol/L$ because the solution already contains $0.10 \,mol / L$ of $OH ^{-}$ from $NaOH$

$K_{ sp }=2.0 \times 10^{-15}=\left[ Ni ^{2+}\right]\left[ OH ^{-}\right]^{2}$

$=(S)(0.10+2 S)^{2}$

As $K_{ sp }$ is small, $2 \,S \,<\,<\,0.10$

thus, $(0.10+2 S) \approx 0.10$

Hence,

$2.0 \times 10^{-15}= S (0.10)^{2}$

$S=2.0 \times 10^{-13} \,M =\left[ Ni ^{2+}\right]$