How many kJ are required to completely vaporize a 25.0 gram sample of water at 100°C?

You have to use the latent heat of vaporization of the water.

The latent heat of vaporization of water at 100 °C is approximately 2.26 kJ / g (search this number in some tables; you will find similar numbers aroung 2.23 and 2.26 kJ/g).

Now mutiply 25.0 g * 2.26 kJ / g and you will get 56.5 kJ

The answer is approximately 56.5 kJ.