Answer
$W_{in}=835.18kW$
Work Step by Step
In this case we are working with potential energy:
$W_{p}=mgh=\rho Vgh$
$W_{p}=1050\frac{kg}{m^3}*0.3\frac{m^3}{s}*9.81\frac{m}{s^2}*200m=618.03kW$
As we have an efficiency, the input power we need:
$W_{in}=\frac{618.03kW}{0.74}=835.18kW$
