The molar solubility of $Cd{\left(OH\right)}_2$ is $1.84\times {10}^{-5}$ M in water.  The expected solubility of $Cd{\left(OH\right)}_2$  in a buffer solution of pH = 12 is 

Given molar solubility,  $s=1.84\times {10}^{-5}$
$K_{sp}=4s^3=4{\left(1.84\times {10}^{-5}\right)}^3$
 $Cd{\left(OH\right)}_2\rightleftharpoons C{d}^{2+}+2O{H}^{-}$
  $s\text{'}$ represent the solubility in buffer solution 
pH = 12,  pOH = 2
 $\left[O{H}^{-}\right]={10}^{-2}$
So,  $s\text{'}=2.492\times {10}^{-10}moles{L}^{-1}$