Understand electricity and magnetism,well past the formulas.

Answer

Full solution

1. Do not add four vectors if you can avoid it. Replace the $-q$ in the lower-left corner by $+q$ and $-2q$ together. That is legal: $+q+(-2q)=-q$.

2. Now the square has $+q$ at all four corners. Opposite corners push on the center in opposite directions with equal strength, so those four contributions cancel exactly.

3. All that remains is the extra $-2q$ at the lower-left corner. The center sits on the diagonal at distance $r=\sqrt{a^2+a^2}=a\sqrt2$, so $r^2=2a^2$.

4. Its field magnitude is $E=\dfrac{k(2q)}{2a^2}=\dfrac{kq}{a^2}$. A negative source pulls, so the field at the center points toward that corner.

5. Check by brute force: the upper-right $+q$ pushes toward the lower-left with $\dfrac{kq}{2a^2}$, the lower-left $-q$ pulls toward the lower-left with another $\dfrac{kq}{2a^2}$, and the two remaining $+q$ charges cancel each other. The total is $\dfrac{kq}{a^2}$ toward the lower-left.

Answer

Full solution

1. Find the null point. Between the charges both fields oppose: $\dfrac{4q}{x^2}=\dfrac{q}{(3a-x)^2}$. Taking square roots, $2(3a-x)=x$, so $x=2a$.

2. Along the axis for $0<x<3a$: $E_x(x)=k\left[\dfrac{4q}{x^2}-\dfrac{q}{(3a-x)^2}\right]$. Just left of $2a$ the big charge wins and $E_x>0$; just right of $2a$ the small charge wins and $E_x<0$. The field points toward $x=2a$ from both sides.

3. Slope: $\dfrac{dE_x}{dx}=k\left[-\dfrac{8q}{x^3}-\dfrac{2q}{(3a-x)^3}\right]=k\left[-\dfrac{8q}{8a^3}-\dfrac{2q}{a^3}\right]=-\dfrac{3kq}{a^3}$ at $x=2a$.

4. A charge $Q$ feels $F_x=QE_x$. For $Q<0$ the force is $F_x\approx +\dfrac{3kq|Q|}{a^3}(x-2a)$: it points the same way as the displacement, so the charge runs away along the axis. That already makes the statement false.

5. For completeness, the sideways direction goes the other way. At $(2a,y)$ with small $y$, both charges push a positive test charge away from the axis: $E_y\approx k\left[\dfrac{4q\,y}{(2a)^3}+\dfrac{q\,y}{a^3}\right]=\dfrac{3kq}{2a^3}y$. A negative charge is pulled back toward the axis, so it is stable sideways but unstable along the axis. No sign of charge is stable in all directions at once, because the field has no source at that empty point, so what spreads out along one axis must converge along the others.