Answer
$0.37$
Work Step by Step
This problem is just the reverse of the previous one.
We know that the thermal efficiency of this power plant is given by
$$\eta=\dfrac{W_{out}}{Q_H}$$
where $W_{out}=P_{out}t$ where $P$ is the power and $t$ is the time, $Q_H=W_{out}+Q_C$.
Thus,
$$\eta=\dfrac{P_{out}t}{P_{out}t+Q_C}$$
where $Q_C$ is the heat absorbed by the water and is given by
$$Q_C=mc\Delta T$$;
$$\eta=\dfrac{P_{out}t}{P_{out}t+mc\Delta T}$$
Dividing the numerator and denominator of the right side by $1/t$.
$$\eta=\dfrac{P_{out} }{P_{out} +\left[\dfrac{m}{t}\right]c\Delta T}$$
where $m=\rho V$;
$$\eta=\dfrac{P_{out} }{P_{out} +\left[\dfrac{V}{t}\right]\rho c\Delta T}$$
Plugging the known;
$$\eta=\dfrac{(750\times 10^6) }{(750\times 10^6) +\left[\dfrac{1\times 10^8\times 10^{-3}}{1\times 60\times60}\right](1000)(4190)(27-16)}$$
$$\eta=\color{red}{\bf 0.37}$$
