Answer
a) $48\;\rm m$
b) $0.32$
Work Step by Step
a) The volume of the coal needed for one day is given by
$$\rho=\dfrac{m}{V}$$
Hence,
$$V=Ah=\dfrac{m}{\rho} $$
where $A$ is the cross-sectional area and $h$ is the height.
Hence, the height needed is given by
$$ h=\dfrac{m}{A\rho} $$
We know that 1 metric ton is 1000 kg, so the total mass is given by multiplying the mass needed for one hour by the 24 hours of one day.
$$m=300\times 1000\times 24$$
Plugging into (1) and plug the given;
$$h=\dfrac{300\times 1000\times 24}{10\times 10\times 1500} $$
$$h=\color{red}{\bf 48}\;\rm m$$
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b) To find the power plant's thermal efficiency, we can treat it as a heat engine.
$$\eta=\dfrac{W_{out}}{Q_H}$$
where $W_{out}=P\Delta t$ where $P$ is the power and $\Delta t$ is the time.
$$\eta=\dfrac{P\Delta t}{Q_H}$$
And we know that the planet generates 750 Mw per hour which is the output work.
The total heat due to the coal burning for one hour is given by
$$Q_H=m\times {\text{ heat of combustion per 1 kg}}$$
Hence,
$$\eta=\dfrac{P\Delta t}{m\times {\text{ heat of combustion per 1 kg}}}$$
Plugging the known;
$$\eta=\dfrac{(750\times 10^6)(60^2)}{(300\times 10^3)(28\times 10^6)}$$
$$\eta=\color{red}{\bf 0.32}$$
