The correct option of this question will be** (B).**

**Solution —**

Here,

Mass of water, $m=1 \mathrm{~kg}$

Latent heat of ice, $L=336 \mathrm{~kJ} \mathrm{~kg}^{-1}$

Temperature of hot reservoir (i.e. room),

$T_{1}=24.4^{\circ} \mathrm{C}=24.4+273=297.4 \mathrm{~K}$

Temperature of cold reservoir(i.e. water),

$T_{2}=0^{\circ} \mathrm{C}=0+273=273 \mathrm{~K}$

The amount of heat extracted from water at $0^{\circ} \mathrm{C}$ to convert to ice at $0^{\circ} \mathrm{C}$ is

$Q_{2}=m L=(1 \mathrm{~kg})\left(336 \mathrm{~kJ} \mathrm{~kg}^{-1}\right)=336 \mathrm{~kJ}$

The coefficient of performance $(\alpha)$ of a refrigerator is

$\alpha=\frac{Q_{2}}{W}=\frac{T_{2}}{T_{1}-T_{2}} \quad \therefore \quad W=\frac{Q_{2}\left(T_{1}-T_{2}\right)}{T_{2}}$

or $W=\frac{(336 \mathrm{~kJ})(297.4 \mathrm{~K}-273 \mathrm{~K})}{273 \mathrm{~K}}$

$=\frac{(336 \mathrm{~kJ})(24.4 \mathrm{~K})}{273 \mathrm{~K}}=30 \mathrm{~kJ}$