0.35 kg of ice at 0 °C melts, and the water then warms to 20 °C…

Step 1. Melting. $\dfrac{0.35\ \text{kg} \times 3.34 \times 10^{5}\ \text{J/kg}}{273.15\ \text{K}} = 428\ \text{J/K}$.

Step 2. Warming. $0.35\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times \ln\dfrac{293.15\ \text{K}}{273.15\ \text{K}} = 103.5\ \text{J/K}$.

Step 3. In all. $428.0\ \text{J/K} + 103.5\ \text{J/K} = 531.5\ \text{J/K}$.