Answer
$$5.460393400 \times 10^{11} \mathrm{N}$$
Work Step by Step
Let $y=0$ be at the bottom of the dam, so that the top of the dam is at $y=185 .$ Then the width of the dam at height $y$ is $$2000+\frac{1000 y}{185} .$$ The dam is inclined at an angle of $55^{\circ}$ to the horizontal, so the height of a horizontal strip is
$$
\frac{\Delta y}{\sin 55^{\circ}} \approx 1.221 \Delta y
$$
so that the area of such a strip is
$$
1.221\left(2000+\frac{1000 y}{185}\right) \Delta y
$$
Then
\begin{aligned}
F &=\rho g \int_{0}^{185} 1.221 y\left(2000+\frac{1000 y}{185}\right) d y\\
&=\rho g \int_{0}^{185} 2442 y+6.6 y^{2} d y\\
&=\left.\rho g\left(1221 y^{2}+2.2 y^{3}\right)\right|_{0} ^{185} \\
&=55,718,300 \rho g\\
&=55,718,300 \cdot 9800\\
&=5.460393400 \times 10^{11} \mathrm{N}
\end{aligned}
