Solution

In a compressed fluid, the pressure $p$ is uniform throughout the volume. It is a known fundamental principle that the vector sum of the pressure forces acting from the fluid on any closed surface is zero.

Let the force vectors acting on the four faces of the tetrahedron be $\vec{F}_1, \vec{F}_2, \vec{F}_3$, and $\vec{F}_4$. The equilibrium condition is:
$$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = 0$$

According to the problem, the resultant force acting on the three upper faces is $\vec{F}$:
$$\vec{F}_{123} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3$$
The magnitude of this resultant force is $|\vec{F}_{123}| = F$.

Therefore, the force acting on the fourth (bottom) face must completely balance this resultant force:
$$\vec{F}_{123} + \vec{F}_4 = 0 \implies \vec{F}_4 = -\vec{F}_{123}$$

From this, it follows that the magnitude of the pressure force on the fourth face is also equal to $F$:
$$|\vec{F}_4| = F$$

By the definition of pressure, the magnitude of the force acting on a flat face is equal to the product of the pressure $p$ and the area of this face $S$:
$$F_4 = pS \implies pS = F$$

Since the tetrahedron is regular, all its faces are identical equilateral triangles with a side length of $a$. The area of one such face is calculated by the formula:
$$S = \frac{a^2 \sqrt{3}}{4}$$

Substituting the area into the force equation, we obtain:
$$p \cdot \frac{a^2 \sqrt{3}}{4} = F$$

Solving for the required fluid pressure, we get:
$$p = \frac{4F}{a^2 \sqrt{3}}$$