pH of saturated solution of $Ba{\left(OH\right)}_2$ is 12.  The value of solubility product Ksp of $Ba{\left(OH\right)}_2$ is 

$\left[O{H}^{-}\right]={10}^{-2}$
$Ba{\left(OH\right)}_2\left(s\right)\rightleftharpoons B{a}^{2+}+2O{H}^{-}$
      s                       s            2s
$Ksp=\left[B{a}^{2+}\right]{\left[O{H}^{-}\right]}^2=\frac{\left({10}^{-2}\right)}{2}\times {\left({10}^{-2}\right)}^2$
$=5.0\times {10}^{-7}$