Answer
Greater than.
Work Step by Step
We know that the electric flux is given by
$$\Phi=EA\cos\theta$$
And in both cases here $\theta=0^\circ$.
Thus,
$$\Phi=EA $$
Since the electric field is uniform and constant for both objects and since the area of the square is greater than the area of the circle, the electric flux of the square is greater than that of the circle.
Thus,
$$\boxed{\Phi_{\rm square}\gt \Phi_{\rm circle}}$$
