Using equation of motion, $v^2=u^2+2as$

$v^2=0+2\left(-g\right)\left(-h\right)$

$\Rightarrow v^2=2gh$

Initial velocity of ball is $v=\sqrt{2gh}=\sqrt{20h}$.

 If after entering the water this velocity does not change, then this value should be equal to the terminal velocity.

Then, $\sqrt{2gh}=\frac{2}{9}\frac{r^2\left(\rho -\sigma \right)g}{\eta }$, here, $\rho$ is density of ball, $\sigma$ is density of water and $\eta$ is viscosity.

$$\begin{gathered} \Rightarrow \sqrt{20h}=\frac{2}{9}\frac{{\left(0.1\times {10}^{-3}\right)}^2\left({10}^4-{10}^3\right)10}{1\times {10}^{-5}} \\ \Rightarrow \sqrt{20h}=\frac{2}{9}\frac{{10}^{-8}\left(10-1\right){10}^3\times 10}{1\times {10}^{-5}} \end{gathered}$$

$\Rightarrow 20h=400$

$\therefore h=20 \mathrm{m}$