This is physics, not math.
If a mass $m$ of water falls through a height $h$ then the potential energy change is $mgh$.
If $\dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $\dot{m}gh$.
If water has an average density of $\rho$, and the volume flow rate is $\dot{V}$ then the mass flow rate is $\dot{m} = \rho \dot{V}$.
If all the potential energy is converted to usable work $P$ in the turbine then we have $P = \rho \dot{V} gh$.
You have $P,h,g,\rho$. Compute $\dot{V}$.
**Addendum** :
$\rho$ is the density of water, we usually take $\rho = 1000\ kg/m^3$.
$g$ is the acceleration due to gravity, we usually take $g = 9.81 \ m/s^2$.
In the above, $h=3 \ m$, $P= 50 \times 10^6 \ W$.
The dot as in $\dot{m}$ usually means the rate of change with respect to time. So, $\dot{V}$ is the change of volume per unit time, or flow rate.