Mass of ice, m = 1kg at -10°C 
mass of water, M = 4.4kg at 30°C 
heat flows through water to ice. so, heat gained by ice and heat lost by water. 
total heat gained by ice to melt = heat required by ice to reach the temperature 0°C + heat require to melt of ice 
= ms∆T + mL 
= 1 kg × 2100 J/Kg/°C × (0 -(-10)) + 1kg × 336000/Kg
= 21000 + 336000 
= 357000 J 
heat lost by water to reduce the temperature at 0°C = MS∆T'
= 4.4kg × 4200 × (30 - 0) = 554400> 357000 
so, temperature of mixture must be greater than zero. 
now, let's take, after melting ice reaches temperature A. 
then, heat is required to change temperature A is (554400 - 357000) = 197400J 
use formula, (m + M)S∆T = 197400 
or, (1 + 4.4)× 4200 × (A - 0) = 197400
A = 8.7°C
 
						

							
Dear student 
Check out this short method.
Given that,
Mass of ice = 1 kg
Temperature of ice = 283 K
Mass of water = 4.4 kg
Temperature of water = 303 K
Specific heat of ice = 2100 J/kg-K
We know that,
Specific heat of water = 4185.5 J/kg-K
Now, the final temperature of mixture is :
Tf = miTiCpi + mwTwCpw /(miCpi + mwCpw)
Put al the values 
Tf = 301k 
Hope this helps