Answer
15 kg.
Work Step by Step
Use the given work input, and the Carnot efficiency, to find the energy that is removed from the low-temperature reservoir.
$$e=1-\frac{T_L}{T_H}=\frac{W}{Q_H}=\frac{W}{W+Q_L}$$
$$Q_L=W\frac{T_L}{T_H-T_L}$$
Now, following the hint in the problem, use the latent heat of vaporization to determine the mass of water.
$$Q_L=W\frac{T_L}{T_H-T_L}=mL_V$$
The heat of vaporization is given on page 399 in Chapter 14 for evaporation at $20^{\circ}C$.
$$m=\frac{W}{L_V}\frac{T_L}{T_H-T_L}$$
$$m=\frac{(600W)(3600s)}{2.45\times10^6 J/kg}\frac{(273+8)K}{17K}=14.6kg\approx 15kg$$