The idea here is that you need to take into account the heat that you must supply to the *ice* at #0^@"C"# to get it to *water* at #0^@"C"#, then to the water at #0^@"C"# to get it water at #25^@"C"#.

An important thing to notice here is that the *enthalpy of fusion* is given to you in kilojoules **per mole**. You can convert this to kJ **per gram** to make the rest of the calculations easier by using water's molar mass

#6.01"kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/("18.02 g") = "0.335 kJ/g"#

Now, to get the water to undergo a phase change, i.e. make it go from solid to liquid at *constant temperature*, #0^@"C"#, you need to supply

#q_1 = m_"water" * DeltaH_"fus"#

#q_1 = 36color(red)(cancel(color(black)("g"))) * 0.335"kJ"/color(red)(cancel(color(black)("g"))) = "12.06 kJ"#

To heat the water from liquid at #0^@"C"# to liquid at #25^@"C"#, you need to use the equation

#q_2 = m_"water" * c_"water" * DeltaT#, where

#c_"water"# - the specific heat of water;

#DeltaT# - the change in temperature, calculated at the final temperature minus the initial temperature.

In your case, you have

#DeltaT = 25 - 0 = 25^@"C"#

This is equal to #"25 K"#, since you have

#DeltaT = (25 + 273.15) - (0 + 273.15)#

#DeltaT = 25 + color(red)(cancel(color(black)(273.15))) - 0 - color(red)(cancel(color(black)(273.15))) = "25 K"#

Therefore, you get

#q_2 = 36color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)("K")))) * (25 - 0)color(red)(cancel(color(black)("K")))#

#q_2 = "3762 J" = "3.762 kJ"#

The *total* heat needed will be

#q_"total" = q_1 + q_2#

#q_"total" = 12.06 + 3.762 = color(green)("15.8 kJ")#

I'll leave the answer rounded to three sig figs, despite the fact that you only gave two sig figs for the mass of the water.