An ice cube of volume 6.1 cm3 is initially at a temperature of -13.6°C. How much heat is required to convert this ice cube into steam?

The specific heat of water is approximately

specific\,heat=\frac {1\,cal}{gram\,deg\,C}

it's close to that for ice, but I'm not exactly sure how close.

The heat required is
1. The specific heat of ice times the temperature it has to be raised to 0 C.
2. The specific heat of water times the temperature difference to 100 C (100 C),
3. Add the latent heat of fusion multiplied by the weight (the heat required to freeze or melt ice)
4. Add the latent heat of vaporization multiplied by the weight (the heat required to evaporate water or condense steam to water).

Add it all together and you have your answer.