Problem with the solubility and Ksp of zinc hydroxide
There are no simple relation between the solubility and the solubility product when a doubly charged ion is involved. The measured solubility is always much bigger than the value obtained from the solubility product. This is due to the doubly charged ion. 

Doubly charged ions like $\ce{Zn^{2+}}$ are usually hydrolyzed in water, and are partly transformed into basic ions like $\ce{[Zn(OH)]^+}$ by the following reaction : $$\ce{Zn^{2+} + H2O -&gt; [Zn(OH)]^+ + H+}$$ If $\ce{Zn(OH)_2}$ is dissolved in water, it will not only produce $\ce{Zn^{2+}}$ ions but also basic ions by the following equation :
$$\ce{Zn(OH)_2 -&gt; [Zn(OH)]^+ + OH^-}$$
As a consequence, the concentration of non complexed $\ce{Zn^{2+}}$ may be calculated from the solubility product of $\ce{Zn(OH)_2}$ but this $\ce{[Zn^{2+}]}$ is much smaller than the measured solubility of $\ce{Zn(OH)_2}$, because the measured solubility of $\ce{Zn(OH)_2}$ is the sum of $\ce{[Zn^{2+}]}$ and $\ce{[Zn(OH)]^+}$. It is even possible that ion pairs like  $\ce{Zn^{2+}(OH^-)_2}$ are dissolved in water without being dissociated. 

In the case of $\ce{Zn(OH)_2}$, the measured solubility is $4.2 ·10^{-6}$ M. And the concentration of $\ce{Zn^{2+}}$ taken from $\ce{K_{sp} = 3·10^{-16}}$ is $\ce{[Zn^{2+}] = 1.95·10^{-6}}$ M. This is much less than the measured solubility. It means that the total concentration of the dissolved ion pairs $\ce{Zn^{2+}(OH^-)_2}$ and of the basic ions $\ce{Zn(OH)^+}$ is equal to the difference $(4.2 - 1.95)·10^{-6} = 2.25·10^{-6}$ M. Only $46.4$% of all the dissolved $\ce{Zn(OH)_2}$ produce  simple $\ce{Zn^{2+}}$ ions (which may be $\ce{[Zn(H_2O)_6]^{2+}}$ by the way).