Answer
$x = 6.0~cm$
Work Step by Step
We can write a general equation for the net electric potential:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
Note that $q_2 = -3q_1$
We can write an expression for the net electric potential at a point due to these two charges:
$V_{net} = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$
$V_{net} = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1}{r_1}-\frac{3q_1}{r_2})$
If $(\frac{q_1}{r_1}-\frac{3q_1}{r_2}) = 0$, then $V_{net} = 0$
If $r_2 = 3r_1$, then $V_{net} = 0$
Suppose $x$ is a positive value of $x$ where $V_{net} = 0$
Then $r_1 = x$ and $r_2 = 24.0-x$
We can find $x$:
$r_2 = 3r_1$
$24.0-x = 3x$
$4x = 24.0$
$x = 6.0~cm$
