#### Answer

The third charge has magnitude $\frac{4q}{9}$, a negative charge, and position $x = \frac{L}{3}$

#### Work Step by Step

We can write Coulomb's Law:
$F = \frac{k~q_1~q_2}{r^2}$
Since both charges are positive, the force would push the two charges away from each other. To create equilibrium, the third charge must be negative.
Since the force is proportional to $\frac{1}{r^2}$ the distance to the $4q$ charge must be double the distance to the $q$ charge.
Then the position of the third charge must be $\frac{L}{3}$
We can find the magnitude of $q_3$:
$F = \frac{k~q~q_3}{(L/3)^2} = \frac{k~(q)~(4q)}{L^2}$
$9q_3 = 4~q$
$q_3 = \frac{4q}{9}$
The third charge has magnitude $\frac{4q}{9}$, a negative charge, and position $x = \frac{L}{3}$