Answer: 2.4 K

Given, Heat supplied to 100 gm of water, Q = 1 kJ = 1000J

mass of water, m =100 g = $\frac{100 g}{18 gmol^{-1}}$ = 5.55 mol

Heat supplied to m gm of water, $Q = mC\Delta t$

$\Delta t = \frac{Q}{mC} = \frac{1000J}{\text{5.55 mol} \times 75 JK^{-1}mol^{-1}}$ = 2.4 K