Answer
$m_{evaporated}=7.93*10^-4\frac{kg}{s}$
Work Step by Step
The rate of heat transferred to the water is:
$Q_{transferred}=0.6*3kW=1.8kW=1.8\frac{kJ}{s}$
At a temperature of $95^{\circ}C$ we need $2269.6kJ$ to vaporize $1kg$ of saturated liquid water:
$m_{evaporated}=\frac{1.8\frac{kJ}{s}}{2269.6\frac{kJ}{kg}}=7.93*10^-4\frac{kg}{s}$
