PHY...............

This question has nothing to do with latent heat of vaporization at all.
Rephrase:Specific latent heat of fusion of ice is 3.34x10^5 J kg-1
Specific heat capacity of water = 4200Jkg-1℃-1should be given.

Energy needed to change 0.2 kg of ice at 0℃ to water at 50℃
=Energy needed to change 0.2 kg of ice at 0℃ to water at 0℃+Energy needed to change 0.2 kg of water at 0℃ to water at 50℃

Energy needed to change 0.2 kg of ice at 0℃ to of water at 0℃
= 0.2 kg (3.34x105J kg-1)
= 6.68 x 104J

Energy needed to change 0.2 kg of water at 0℃ to water at 50℃
= 0.2 kg (4200Jkg-1℃-1)(50℃-0℃)
= 4.2 x 104J

Energy needed to change 0.2 kg of ice at 0℃ to water at 50℃
= 6.68 x 104J + 4.2 x 104J
=1.088 x 105J

first,calculate the 0'C of ice change to 0'C water:0.2x3.34x10^5 kg-1.
then,calculate 0'C of water to 50'C water:0.2x2.26x10^6 J kg-1x(50-0)
at last,please put the first step and second step of sum,there is an answer.