Answer
The molar solubility of $Cd(OH)_2$ is $1.8*10^{-5}$ M.
Work Step by Step
Suppose $x$ (M) is the molar solubility of $Cd(OH)_2$ in water. We can construct the following table:
$Cd(OH)_2 (s)$ $ Cd^{2+} (aq) + 2OH^{-} (aq)$
Initial: $[Cd(OH)_2]=x, [Cd^{2+}]=0, [OH^{-}]=0$ (we can neglect the amount of $OH^{-}$ formed by the hydrolysis of water)
Equilibrium: $[Cd(OH)_2]=0, [Cd^{2+}]=x, [OH^{-}]=2x$
Now, we can write the solubility-product expression. Because the term "molar solubility" describes a substance's concentration in a saturated solution, we know that the product is equal to $Ksp$.
$Ksp=[Cd^{2+}][OH^{-}]^2=2.5*10^{-14}$
$x*(2x)^2=2.5*10^{-14}$
$4x^3=2.5*10^{-14}$
$x=1.8*10^{-5}$ M
