Answer
$COP_{}=0.20$
Work Step by Step
The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks, $$
\begin{aligned}
m & =\rho V=9 \times(1 \mathrm{~kg} / \mathrm{L})(0.350 \mathrm{~L})=3.15 \mathrm{~kg} \\
Q_{\text {cosling }} & =m c \Delta T=(3.15 \mathrm{~kg})\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(25-3)^{\circ} \mathrm{C}=290 \mathrm{~kJ} \\
\dot{Q}_{\text {cosling }} & =\frac{Q_{\text {cooling }}}{\Delta t}=\frac{290 \mathrm{~kJ}}{12 \times 3600 \mathrm{~s}}=0.00671 \mathrm{~kW}=6.71 \mathrm{~W}
\end{aligned}
$$ The electric power consumed by the refrigerator is $$
\dot{W}_{\text {in }}=\mathrm{V} I=(12 \mathrm{~V})(3 \mathrm{~A})=36 \mathrm{~W}
$$ Then the COP of the refrigerator becomes $$
\mathrm{COP}=\frac{\dot{Q}_{\text {owoling }}}{\dot{W}_{\text {in }}}=\frac{6.71 \mathrm{~W}}{36 \mathrm{~W}}=\mathbf{0 . 1 8 6}=0.20
$$
