Square Zero Field

Take any point within the square. Only along the vertical and horizontal lines of symmetry does there seem to be a cancellation of fields of some kind, by inspection. Therefore, the point of interest must be on one of these lines and there are in total 4 such points. This calculation focusses on finding one such point. The other three can be found by symmetry. One can also consider points along the diagonal. There may be points where the net field cancels out. However, this has not been checked by me.

Consider the point $P$ lying on the line of symmetry of the square as seen in the figure. Let $\angle PDC = \theta$ . Make a line segment connecting points A and P and let $\angle PAB = \phi$

The horizontal components of the fields cancel out due to symmetry. The vertical components need to be worked out. Assume:

$\frac{q}{4\pi\epsilon_o} = 1$

Let the distance between the midpoint of the line $DC$ and $P$ be $h$ . Therefore:

$h = 0.5\tan{\theta}$ $1-h = 0.5\tan{\phi}$

From here:

$\tan{\theta} + \tan{\phi} = 2 \qquad \dots(1)$

Now, balancing the vertical components of electric fields at point $P$ gives the relation:

$\sin{\theta}\cos^2{\theta} = \sin{\phi}\cos^2{\phi} \qquad \dots(2)$

So solving equations $(1)$ and $(2)$ numerically gives:

$\theta = 0.2228 \ rad$

Another solution is obviously $\theta = \pi/4$ . The distance between the center and this point is, therefore:

$\boxed{D = 0.5(1 - \tan{\theta}) = 0.3867}$

I have skipped a few steps here. If necessary, I will elaborate later.