Answer
$m_{evaporated}=7.98*10^-4\frac{kg}{s}$
Work Step by Step
The rate of heat transferred is:
$Q_{transferred}=0.6*3kW=1.8kW=1.8\frac{kJ}{s}$
As you need $2256.4kJ$ to vaporize $1kg$ of saturated liquid water:
$m_{evaporated}=\frac{1.8\frac{kJ}{s}}{2256.4kJ}=7.98*10^-4\frac{kg}{s}$
