Answer
A greater force is experienced in the corner of the square (see step-by-step for details).
Work Step by Step
The magnitude of the force between two point charges, $q_{1}$ and $q_{2}$, separated by a distance $r$ is
$F =k\displaystyle \frac{|q_{1}||q_{2}|}{r^{2}},\qquad (19- 5)$
(Force is inversely proportional to the square of the distance between charges.)
Let the square have sides a. Then the diagonal is $d=a\sqrt{2}$.
The center of the square is at a distance of about $0.7a$ from the corners.
Placing the third charge in the center, two forces will act on it, along the diagonal. If the third charge is positive, each force is a repelling force, acting in opposite directions. The net force has magnitude zero.
If the third charge is negative, the forces are attracting, but still act in opposite direction, yielding net force zero.
Placing the third charge in a corner, it would be at a greater distance from either charge, so the forces acting would be weaker than above.
But, this time their directions are perpendicular, so they do not cancel each other. A nonzero net force would exist.
Conclusion: greater force is experienced in the corner of the square.
