How many minutes? (Calorimetry)

A design of an electric heater is rated with a power of 15kW. If 10kg of ice at 200K is placed in the heater, how many minutes will it take to change ice to steam at 400K. Assume 100% efficient.

You should have specified the following values:

specific heat capacity of ice = 2,108 J/kg K
specific heat capacity of water = 4,187 J/kg K
specific heat capacity of steam = 1,996 J/kg K
specific latent heat of fusion of ice = 334,000 J/kg
specific latent heat of vaporisation of water = 2,257,000 J/kg
melting point of ice = 273 K
boiling point of water = 373 K

I'm going to use the above values because you didn't provide them.
I'm going to use the two equations below:
energy required to raise temp = mass * specific heat capacity * change in temp
energy required to change state = mass * specific latent heat of fusion/vaporisation

First, the ice which is at 200 K will get warm until it reaches 273 K. After that, it will start melting and at that time, there will be no change in temp. After melting has finished, water will be formed and the water will get hotter and hotter until it reaches 373 K. After that, the water will begin boiling and at that time, there will be no change in temp. After boiling has finished, steam will form and the steam will be heated more until it reaches a temp of 400 K

Energy required to raise the temperature of ice from 200 K to 273 K = 10 * 2,108 * (273 - 200)
= 1,538,840 J
Energy required to melt the ice at 273 K = 334,000 * 10
= 3,340,000 J
Energy required to raise the temperature of water from 273 K to 373K = 10 * 4,187 * (373 - 273)
= 4,187,000 J
Energy required to boil the water at 373 K = 2,257,000 * 10