Answer
See the detailed answer below.
Work Step by Step
a) The maximum possible thermal efficiency of this power plant is given by Carnot efficiency.
$$\eta_{\rm Carnot}=1-\dfrac{T_C}{T_H}$$
Plugging the known;
$$\eta_C=1-\dfrac{(25+273)}{(300+273)}=\color{red}{\bf 0.48}$$
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b) The planet's actual efficiency is given by
$$\eta=\dfrac{W_{out}}{Q_H}=\dfrac{P_{out}\color{red}{\bf\not} t}{P_{in}\color{red}{\bf\not} t}$$
where $P$ is the power and $t$ is the time.
Recalling that the power is given by energy divided by time.
Plugging the known;
$$\eta= \dfrac{1000\times 10^6}{3000\times 10^6}=\color{red}{\bf 0.33}$$
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c) The heat lost in water $Q_C$ to cool down the reactor is given by
$$Q_H=Q_C+W_{out}$$
Hence,
$$Q_C=Q_H-W_{out}$$
which is also given by $mc_{water}\Delta T$, so that
$$Q_C=P_{in}t-P_{out}t=mc_{water}(T_f-T_i)$$
Thus,
$$T_f=\dfrac{P_{in}t-P_{out}t}{mc_{water}}+T_i$$
$$T_f=\dfrac{(P_{in} -P_{out})t}{mc_{water}}+T_i$$
Dividing the numerator and denominator of the first term on the right side by $1/t$.
$$T_f=\dfrac{P_{in} -P_{out} }{\left(\dfrac{m}{t}\right)c_{water}}+T_i$$
where $m=\rho V$
$$T_f=\dfrac{P_{in} -P_{out} }{\left(\dfrac{ V}{t}\right)\rho c_{water}}+T_i$$
Plugging the known in SI units only:
$$T_f=\dfrac{(3000\times 10^6) -(1000\times 10^6) }{\left(\dfrac{ 1.2\times 10^8\times 10^{-3}}{1\times 60\times 60}\right)(1000)(4190)}+(18+273)\\
T_f=\bf305\;\rm K$$
For water-specific heat, see Table 17.2.
Thus the water exit at a temperature of
$$T_f=\color{red}{\bf 32}\rm ^\circ C$$
