Start with the definition of power. Replace work with potential energy. We will ignore the kinetic energy of the wheat, since our real goal is to elevate the grain, not accelerate it. Replace mass with density times volume.
Rearrange it just a bit and substitute the numbers given to compute the power delivered as useful work.
|P = ρ||V||gh = (770 kg/m3)||(900 m3)||(9.8 m/s2)(60 m) = 113 kW|
Compare this to the rated power of the motor to get the efficiency of the grain elevator.
|η =||Wout||=||Pout||=||113 kW||= 75%|