Answer
$2.4\times 10^6 \;\rm {L}/{min}$
Work Step by Step
The waste heat power is about 2000 MW as the author told us.
And we know that this loss of power goes to heat the water.
We know that the loss-power is given by
$$P_{\rm loss}=\dfrac{-Q_{reactor}}{t}$$
where $Q_{reactor}$ is heat lost from the reactor to the water and $t$ is the time. Noting that the negative sign is due to losing the heat.
Hence,
$$Q_{reactor}=-P_{\rm loss}t\tag 1$$
And we know that the heat gained by the water is given by
$$Q_{water}=m_{water}c_{water}\Delta T\tag 2$$
If the system is an isolated system,
$$Q_{reactor}+Q_{water}=0$$
Plugging from (1) and (2);
$$-P_{\rm loss}t+m_{water}c_{water}\Delta T=0$$
$$ P_{\rm loss}t=m_{water}c_{water}\Delta T $$
The author asks about the volume of water, so we need to recall the density law of $m =\rho V$,
$$ P_{\rm loss}t=\rho_{water}V_{water}c_{water}\Delta T $$
Solving for $V$;
$$V_{water}=\dfrac{P_{\rm loss}t}{\rho_{water} c_{water}\Delta T }$$
Plugging the known;
$$V_{water}=\dfrac{(2000\times 10^6)(60)}{(1000)(4190)(30-18) }$$
$$V_{water}=2386\;\rm m^3$$
This is the amount of water heated in 60 seconds or 1 minute.
The volume of litter in one minute is then given by
$$\dfrac{V_{water}}{t} \approx \color{red}{\bf2.4\times 10^6}\;\rm \frac{L}{min}$$
