As we know that the Carnot engine is the most efficient heat engine operating between two temperature $T_1$ and $T_2$. Refrigerator is also Carnot's engine working in reverse order. Its efficiency is $\eta$. The efficiency of Carnot engine is $\eta=1-\frac{T_2}{T_1}$
As given that temperature of the soruce is $27^{\circ} \mathrm{C}$ so, $T_1=(27+273) \mathrm{K}=300 \mathrm{~K}$
Temperature of sink is $\left(-3^{\circ} \mathrm{C}\right)$ so, $T_2=(-3+273) \mathrm{K}=270 \mathrm{~K}$
Efficiency of a perfect heat engine is given by $\eta=1-\frac{T_2}{T_1}=1-\frac{270}{300}=\frac{1}{10}$
Efficiency of refrigerator is $50 \%$ of a perfect engine. Efficiency of refrigerator $=50 \%$ of $1=0.5$.
$$
\eta^{\prime}=0.5 \times \eta=\frac{1}{2} \eta=\frac{1}{20}=0.05
$$
Coefficient of performance of the refrigerator $(\beta)$
$$
\begin{aligned}
&=\frac{Q_2}{W}=\frac{1-\eta^{\prime}}{\eta^{\prime}}=\frac{1-0.05}{0.05}=\frac{0.95}{0.05}=19 \\
&\begin{array}{l}
Q_2=\beta W \\
\quad=19 \% \text { work done by motor on refrigerator } \\
\quad=19 \times(1 \mathrm{~kW})=19 \mathrm{~kW}=19 \mathrm{~kJ} / \mathrm{s} .
\end{array}
\end{aligned}
$$
Hence, heat is taken out of the refrigerator at a rate of $19 \mathrm{~kJ}$ per second.