Answer
$x = 2.00~cm$
Work Step by Step
If particle 3 is placed between particle 1 and particle 2, then the force on particle 3 due to particle 1 is in the +x direction, and the force on particle 3 due to particle 2 is in the +x direction.
The net force on particle 3 is the sum of these two forces.
We can write an expression for the magnitude of the net force in terms of $x$, the x coordinate of particle 3:
$F = \frac{4e^2}{4\pi~\epsilon_0~x^2}+\frac{108e^2}{4\pi~\epsilon_0~(8.00-x)^2}$
We can find the value of $x$ where $F$ is a minimum by finding the derivative:
$\frac{dF}{dx} = \frac{(-2)(4e^2)}{4\pi~\epsilon_0~x^3}+\frac{(2)(108e^2)}{4\pi~\epsilon_0~(8.00-x)^3} = 0$
$\frac{27}{(8.00-x)^3} = \frac{1}{x^3}$
$\frac{3}{8.00-x} = \frac{1}{x}$
$3x = 8.00-x$
$4x = 8.00$
$x = 2.00~cm$
