Answer
a) $\rm 13.2\; km^3/day$
b) $73\;\rm km^2$
Work Step by Step
a)
To find the volume of air that is heated per day, we first need to find the amount of heat delivered to this air.
$$Q=Q_L=m_{air}c_{air}\Delta T_{air}$$
whereas $Q$ is the amount of heat absorbed by air which is the amount of heat exhausted from the cooling towers.
We also know, from the density law, that $m=\rho V$.
So that,
$$Q_L=\rho_{air} V_{air}c_{air}\Delta T_{air}$$
Solving for $V_{air}$;
$$V_{air}=\dfrac{Q_L}{\rho_{air} c_{air}\Delta T_{air}} \tag 1$$
Now we need to find $Q_L$ which we can find from the COP of the power plant.
$$e=\dfrac{W}{Q_H}=\dfrac{W}{W+Q_L}$$
Solving for $Q_L$;
$$W=eW+eQ_L$$
$$Q_L=\dfrac{W(1-e)}{e}$$
Plugging into (1) and then plugging the known;
$$V_{air}=\dfrac{W(1-e)}{e\rho_{air} c_{air}\Delta T_{air}} $$
Divide both sides by $t$;
$$\dfrac{V_{air}}{t}=\dfrac{\dfrac{W}{t}(1-e)}{e\rho_{air} c_{air}\Delta T_{air}} $$
$$\dfrac{V_{air}}{t}=\dfrac{850\times10^6 \cdot (1-0.38)}{0.38\cdot 1.3 \cdot (1\times10^3) \cdot 7 }=1.524\times 10^5\;\rm m^3/s $$
$$\dfrac{V_{air}}{t}=\rm \dfrac{1.524\times 10^5\;\rm m^3}{1\;s}\cdot \dfrac{3600\;s}{1\;h}\cdot \dfrac{24\;h}{1\;day}\cdot \dfrac{1^3\;km^3}{1000^3\;m^3} $$
$$\dfrac{V_{air}}{t}=\color{red}{\bf 13.2}\;\rm km^3/day$$
b)
We know that the volume is given by the area times the height.
$$V_{air}=h_{air}A_{air}$$
whereas $h$ is the thickness of the air layer.
Thus,
$$A_{air}=\dfrac{V_{air}}{h_{air}}=\dfrac{13.2\;\rm km^3}{180\times10^{-3}\;\rm km}=\color{red}{\bf 73.3}\;\rm km^2$$