Answer
78 liters.
Work Step by Step
The coefficient of performance for a refrigerator is given by Eq. 15–6a, $Q_L/W$.
$Q_L$ is the heat removed from the low-temperature area, and is the latent heat removed to freeze the ice. $$COP=\frac{Q_L}{W}=\frac{mL_f}{W}$$
The work done to remove the heat is 1.2 kW multiplied by the elapsed time.
The mass of water frozen is its density multiplied by its volume.
$$COP=\frac{Q_L}{W}=\frac{(\rho V) L_f}{Pt}$$
Solve for the volume.
$$V=\frac{(COP)Pt}{\rho L_f}=\frac{(6.0)(1200W)(3600s)}{(1000kg/m^3)(3.33\times10^5 J/kg)}=0.0778m^3$$