Answer
$$ 5652.37 \mathrm{N}$$
Work Step by Step
Since the fluid surface is at height $y = 1$, the horizontal strip at height y is at a depth of $1 − y.$ Moreover, this strip has a width of $e − e^y$. Thus
\begin{align*}
F&=\rho g \int_{0}^{1}(1-y)\left(e-e^{y}\right) d y\\
&= e\rho g \int_{0}^{1}(1-y) d y-\rho g \int_{0}^{1}(1-y) e^{y} d y\\
&=e\rho g \left.\left(y-\frac{1}{2} y^{2}\right)\right|_{0} ^{1}-\rho g \left.\left((1-y) e^{y}+e^{y}\right)\right|_{0} ^{1}\\
&=\rho g\left(\frac{1}{2} e-(e-2)\right)\\
&=\rho g\left(2-\frac{1}{2} e\right)\\
&=900 \cdot 9.8\left(2-\frac{1}{2} e\right)\\
& \approx 5652.37 \mathrm{N}
\end{align*}
