Answer
Therefore $1.7\times 10^5~m^3$ of water should be pumped to the upper reservoir.
Work Step by Step
The potential energy of the water should be equal to the total energy $E$ produced by the power plant in 1.0 hour. Let $n$ be the number of cubic meters of water.
$PE = E$
$n\times mgh = (180\times 10^6~W)(3600~s)$
$n = \frac{(180\times 10^6~W)(3600~s)}{mgh}$
$n = \frac{(180\times 10^6~W)(3600~s)}{(10^3~kg/m^3)(9.80~m/s^2)(380~m)}$
$n = 1.7\times 10^5~m^3$
Therefore $1.7\times 10^5~m^3$ of water should be pumped to the upper reservoir.
