Answer
$61\;\rm W$
Work Step by Step
The 4800 W that enters the room, when the window is not covered, is the rate of cooling heat. In other words, $\left(\dfrac{Q_L}{t}\right)_1=4800\;\rm W$ for the first case.
And when we cover the window, the cooling rate became,$\left(\dfrac{Q_L}{t}\right)_2=500\;\rm W$.
Now we need to find the power done by the air conditioner in each case and then find the difference between them.
$$e=1-\dfrac{T_L}{T_H}=\dfrac{W}{Q_H}=\dfrac{W}{W+Q_L}$$
Thus,
$$1-\dfrac{T_L}{T_H}=\dfrac{W}{W+Q_L}$$
Solving for $W$;
$$\left[1-\dfrac{T_L}{T_H}\right]W+\left[1-\dfrac{T_L}{T_H}\right]Q_L=W$$
$$W\left(1-\dfrac{T_L}{T_H}-1\right)=-\left[1-\dfrac{T_L}{T_H}\right]Q_L$$
$$W\left( -\dfrac{T_L}{T_H} \right)=-\left[1-\dfrac{T_L}{T_H}\right]Q_L$$
$$W=\dfrac{-\left[1-\dfrac{T_L}{T_H}\right]Q_L}{\left( -\dfrac{T_L}{T_H} \right)}=\dfrac{ \left[1-\dfrac{T_L}{T_H}\right]T_HQ_L}{T_L}$$
$$W=\dfrac{-\left[1-\dfrac{T_L}{T_H}\right]Q_L}{\left( -\dfrac{T_L}{T_H} \right)}=Q_L\cdot\dfrac{T_H-T_L}{T_L}$$
Divide both sides by $t$;
$$\dfrac{W}{t} =\dfrac{Q_L}{t}\cdot\dfrac{T_H-T_L}{T_L}$$
Therefore,
$$P =\dfrac{Q_L}{t}\cdot\dfrac{T_H-T_L}{T_L}$$
Thus,
$$\Delta P=P_1-P_2=\left(\dfrac{Q_{L}}{t}\right)_1\dfrac{T_H-T_L}{T_L}-\left(\dfrac{Q_{L}}{t}\right)_2\dfrac{T_H-T_L}{T_L}$$
$$\Delta P =\dfrac{T_H-T_L}{T_L}\left[\left(\dfrac{Q_{L}}{t}\right)_1-\left(\dfrac{Q_{L}}{t}\right)_2\right]$$
Plugging the known;
$$ \Delta P=\dfrac{(32+273)-(21+273)}{21+273}\left[ 4800-500\right]$$
$$ \Delta P=\color{red}{\bf161}\;\rm W$$
So, the power saved is about 161 W.
