Answer
$F = 4.70\times 10^5~N$
Work Step by Step
When the aquarium is filled to a depth of $4.00~m$, the top $2.00~m$ of water presses on the wall with the same total force as when the water was filled to a depth of $2.00~m$
The additional force comes from the bottom $2.00~m$ of water pressing on the wall.
We can find the average pressure of the bottom $2.00~m$ of water:
$P = \rho~g~h$
$P = (1000~kg/m^3)(9.8~m/s^2)(3.0~m)$
$P = 29,400~Pa$
We can find the additional force of this pressure on the wall:
$F = P~A$
$F = (29,400~Pa)(2.00~m)(8.00~m)$
$F = 4.70\times 10^5~N$
