Answer
$$
F_{\text {water }}=6.8 \times 10^{3} \mathrm{N} .
$$
Work Step by Step
Newton's second law applied to the barge, in the $x$ and $y$ directions, leads to
$$
\begin{aligned}(7900 \mathrm{N}) \cos 18^{\circ}+F_{x} &=m a \\(7900 \mathrm{N}) \sin 18^{\circ}+F_{y} &=0 \end{aligned}
$$
respectively. Plugging in $a=0.12 \mathrm{m} / \mathrm{s}^{2}$ and $m=9500 \mathrm{kg}$ , we obtain $$F_{x}=-6.4 \times 10^{3} \mathrm{N}$$
and $$F_{y}=-2.4 \times 10^{3} \mathrm{N}$$
The magnitude of the force of the water is therefore
$$
F_{\text {water }}=\sqrt{F_{x}^{2}+F_{y}^{2}}=6.8 \times 10^{3} \mathrm{N} .
$$
