Answer
$3\;\rm cm$, $6\;\rm cm$
Work Step by Step
As we see in the figure below, the point at which the net electric potential is zero is at $x$ distance from the origin at which $q_1$ lies.
Hence, $r_1=x$ while $r_2=|x-0.04|$.
Hence,
$$V_{net}=V_1+V_2=0$$
$$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$$
$$\dfrac{q_1}{x}=-\dfrac{q_2}{|x-0.04|} $$
Plug the known and solve for $x$,
$$\dfrac{3}{x}=-\dfrac{-1}{|x-0.04|} $$
Thus,
$$ |x-0.04|=\dfrac{x}{3}$$
Hence,
$$ x-0.04=\pm \dfrac{x}{3}$$
So, we have two points at which the electric potential is zero which are
$$x=\color{red}{\bf 6}\;\rm cm$$
and
$$x=\color{red}{\bf 3}\;\rm cm$$
