Answer
$\eta_{pump}=74.07\%$
Work Step by Step
In this case the mechanic power of the fluid is:
$W_{m,fluid}=Q\Delta P=(50\frac{L}{s}*\frac{1m^3}{1000L})*(300kPa-100kPa)=10kW$
And the mechanic power of the pump is:
$W_{m,pump}=15kW*0.90=13.5kW$
Then the efficiency of the pump is:
$\eta_{pump}=\frac{10kW}{13.5kW}=0.7407$
$\eta_{pump}=74.07\%$
