Answer
The magnitude of the gravitational force is $1.1 \times 10^{-8}~N$ directed toward the center of the square.
Work Step by Step
Let's consider the gravitational forces acting on the sphere in the top left corner of the square. Then the three forces will pull this sphere to the right, straight down, and toward the center of the square.
The force pulling to the right:
$F_{g1} = \frac{Gm^2}{r^2}$
The force pulling straight down:
$F_{g2} = \frac{Gm^2}{r^2}$
The force pulling toward the center of the square:
$F_{g3} = \frac{Gm^2}{(\sqrt{2}~\cdot ~r)^2} = \frac{Gm^2}{2 ~r^2} = 0.5 \times \frac{Gm^2}{r^2}$
The forces acting to the right and straight down will combine to pull the sphere toward the center of the square.
$F_{g12} = \sqrt{(\frac{Gm^2}{r^2})^2 + (\frac{Gm^2}{r^2})^2} = \sqrt{2}\times \frac{Gm^2}{r^2}$
We can add the two forces directed toward the center of the square.
$F_g = F_{g12} + F_{g3} = (\sqrt{2}+0.5)\times \frac{Gm^2}{r^2}$
$F_g = 1.914\times (\frac{(6.67 \times 10^{-11} ~N\cdot m^2/kg^2)(7.5 ~kg)^2}{(0.80 ~m)^2})$
$F_g = 1.1 \times 10^{-8}~N$
The magnitude of the gravitational force is $1.1 \times 10^{-8}~N$ directed toward the center of the square.