Answer
$P_{e}=189.7kW$
Work Step by Step
Here we are working with potential energy:
The potential work is: $W_{p}=mgh$
And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$
The maximum potential power is:
$P_{p}=1000\frac{kg}{m^3}*(0.25\frac{m^3}{s})*9.81\frac{m}{s^2}*85m=208.462kW$
As the efficiency is $91%$, the electric power output is:
$P_{e}=\eta P_{p}=208.462kW*0.91=189.7kW$
