Answer
A charge of $~-0.432~q~$ must be placed at the point $(4.0, 0.0)$.
Work Step by Step
First, we can find the magnitude of the electric field at point $(4.0, 3.0)$ due to the two charges $q$.
By symmetry, the horizontal component of the electric field due to the two charges $q$ is zero.
Let $\theta$ be the angle above the horizontal of the line from each point charge $q$ to the point $(4.0,3.0)$. The vertical component of the electric field at the point $(4.0,3.0)$ is the sum of the vertical components of the electric field due to each point charge $q$:
$E_y = 2\times \frac{k~q}{r^2}~(sin~\theta)$
$E_y = 2\times \frac{k~q}{(5.0~m)^2}~(\frac{3.0}{5.0})$
$E_y = 0.048~k~q$
The third charge at the point $(4.0, 0.0)$ must be negative and the electric field due to this point charge must be equal in magnitude to $E_y$. We can find $Q$, the magnitude of the charge at the point $(4.0, 0.0)$:
$\frac{k~Q}{d^2} = E_y$
$\frac{k~Q}{d^2} = 0.048~k~q$
$Q = 0.048~q~d^2$
$Q = 0.048~q~(3.0~m)^2$
$Q = 0.432~q$
A charge of $~-0.432~q~$ must be placed at the point $(4.0, 0.0)$.
