Flux Squared

For the inverse square field we are dealing with here the flux only depends on the solid angle, and its value is $\phi=\Omega F_0$ , where $F_0=1$ is the field strength at unit distance. We can consider the square as one face of a cube, with the source in the center. Because the full $4\pi$ solid angle is equally divided between 6 faces in our case $\Omega=4\pi/6=2\pi/3$ and the flux is $\phi=\Omega F_0==2\pi/3=2.094$

The flux out of the solid region $\left[-\frac{1}{2},\frac{1}{2}\right]^3$ is $4\pi$ , as with any closed surface enclosing the origin. By symmetry, the flux we seek is a sixth of that, $\frac{2\pi}{3}\approx \boxed{2.094}$ .