Answer
$3v_0$
Work Step by Step
Since the tube that contains the two balls is smooth, then the friction force is negligible and the speeds of the two balls remain constant until they came out of the tube's two ends.
The system here is the two balls and it is an isolated system; so the momentum is conserved.
Before releasing the two balls the initial momentum of the system was zero since the two balls were at rest.
Thus,
$$p_{ix}=p_{fx}\\
p_{fx}=m_1v_{1x}+m_2v_{2x}=0$$
where $m_1=3m$, $v_{1x}=v_0$, and $m_2=m$.
Thus,
$$ 3mv_0 +m v_{2x}=0$$
Hence,
$$ v_{2x}=\dfrac{-3\color{red}{\bf\not} mv_0 }{\color{red}{\bf\not} m}$$
$$ \boxed{v_{2x}=-3v_0}$$
The negative sign here is due to direction, if the heavy ball is released to the right then the other one will move to the left as we got here.
