At 298 k the molar solubility of  $Ca{\left(OH\right)}_2$ in 0.1 M KOH solution is $x\times {10}^{-y}$. The values of x and y are respectively. 
$(At298k,K_{sp}ofCa{\left(OH\right)}_2=2.5\times {10}^{-14})$

We know that,
$\begin{array}{l}\mathrm{Cd}(\mathrm{OH}{)}_2\rightleftharpoons {\mathrm{Cd}}^{2+}+2{\mathrm{OH}}^{-}\\ \mathrm{KOH}=0.1\mathrm{M}=\left[{\mathrm{OH}}^{-}\right]\end{array}$
Let solubility of Cd⁺² ions be S.
$$\begin{gathered} K_{sp}=\left[{\mathrm{Cd}}^{2+}\right]\left[{\mathrm{OH}}^{-}\right] \\ 2.5\times {10}^{-14}=S\times [0.1{]}^2 \\ 2.5\times {10}^{-12}M=S \\ \text{ or }25\times {10}^{-13}M=S \end{gathered}$$
On Comparing the answer with X ×10^-y
We get x = 25 and y = 13.