Answer
$\approx 2.7\;\rm min$
Work Step by Step
Recalling that the total heat energy absorbed by the water is given by
$$Q=mc_w\Delta T\tag 1$$
The author asks about the time it takes to raise its temperature 50$^\circ$C.
So the power of heat transfer is given by
$$P=\dfrac{Q}{t}$$
Hence,
$$t=\dfrac{Q}{P}$$
Plug from (1),
$$t=\dfrac{mc_w\Delta T}{P}\tag 2$$
We know that the energy rate transfer from the beam to the cube is given by
$$P=0.80IA$$
where $I=\frac{1}{2}\epsilon_0cE^2$, and $A=L^2$ where $L$ is the side length of the water cube.
The $0.80$ since the author told us that the water absorbs only 80% of the incident energy.
$$P=\frac{0.80\epsilon_0cE^2L^2}{2}$$
Plug into (2),
$$t=\dfrac{2mc_w\Delta T}{0.80\epsilon_0cE^2L^2} $$
Recall that $m=\rho V$, and $V=L^3$
$$t=\dfrac{2\rho L^3c_w\Delta T}{0.80\epsilon_0cE^2L^2} $$
$$t=\dfrac{2\rho L c_w\Delta T}{0.80\epsilon_0cE^2 } $$
Plug the known;
$$t=\dfrac{2(1000) (0.1) (4186)(50)}{0.80(8.85\times 10^{-12})(3\times 10^8)(11000)^2 } $$
$$t=\color{red}{\bf 163}\;\rm s$$