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Let mass of iron ball = m
At height = 1km =1000m, 
total energy = potential energy = mgh = 1000*9.8*m = 9800m J =9.8m kJ
When all energy is converted into heat and used in raising the temperature of ball,
Specific heat of iron(s) = 0.1 kcal/kg C = 0.42 kJ/kg C 
Let change in temperature = ΔT
ms(ΔT) = mgh
⇒ms(ΔT) = 9.8m
⇒ s(ΔT) = 9.8
⇒ 0.42(ΔT) = 9.8
⇒ ΔT = 9.8/0.42 = 23.3 degree celsius