A closed cubical box is completely filled with water and is...

Consider the water inside the cubical box as the system.

Let the mass of the water be $m$. The box and the water are accelerating horizontally towards the right with an acceleration $a$.

The horizontal forces acting on the water are the normal force from the left wall $N_1$ acting towards the right, and the normal force from the right wall $N_2$ acting towards the left. Since the water is at rest relative to the box, there are no viscous forces acting on it.

Applying Newton's second law of motion in the horizontal direction for the water:

$N_1 - N_2 = ma$

Since the mass $m$ and acceleration $a$ are positive, $ma > 0$.

Therefore, $N_1 - N_2 > 0 \Rightarrow N_1 > N_2$.

By Newton's third law, the force exerted by the water on the left wall is $F_1 = N_1$ and the force exerted by the water on the right wall is $F_2 = N_2$.

Hence, $F_1 > F_2$.

Answer: $F_1 > F_2$