Answer
$ \frac{1}{5}v_0 $
Work Step by Step
The collision is perfectly elastic which means that the energy and the momentum are conserved.
$$E_i=E_f$$
$$K_{i1}+K_{i2}=K_{f1}+K_{f2}$$
where $_1$ refers to the cube and $_2$ refers to the rod.
$$ \color{red}{\bf\not}\frac{1}{2}mv_0^2+0= \color{red}{\bf\not}\frac{1}{2}mv_1^2+ \color{red}{\bf\not}\frac{1}{2}I_2\omega_2^2$$
$$ mv_0^2=mv_1^2+ I_2\omega_2^2$$
where $I_2$ rod is at its center of mass is given by $\frac{1}{12}ML^2$
$$ mv_0^2=mv_1^2+ \frac{1}{12}ML^2\omega_2^2$$
where $L=d$, and $M=2m$;
$$ mv_0^2=mv_1^2+ \frac{1}{12}(2m)d^2\omega_2^2$$
$$ \color{red}{\bf\not}mv_0^2= \color{red}{\bf\not}mv_1^2+ \frac{1}{6} \color{red}{\bf\not}md^2\omega_2^2$$
$$ v_0^2= v_1^2+ \frac{1}{6} d^2\omega_2^2\tag 1$$
And,
$$L_i=L_f$$
The linear momentum just before and just after the collision.
$$mv_0\dfrac{d}{2}=mv_1\dfrac{d}{2}+I_2\omega_2$$
$$\dfrac{1}{2}mv_0d =\dfrac{1}{2} mv_1d +\frac{1}{12}ML^2 \omega_2$$
$$\dfrac{1}{2} \color{red}{\bf\not}mv_0 \color{red}{\bf\not}d =\dfrac{1}{2} \color{red}{\bf\not}mv_1 \color{red}{\bf\not}d +\frac{1}{12}(2 \color{red}{\bf\not}m)d^{ \color{red}{\bf\not}2} \omega_2$$
$$\dfrac{1}{2} v_0=\dfrac{1}{2} v_1+\dfrac{1}{6}d\omega_2 $$
$$ v_0= v_1+\dfrac{1}{3}d\omega_2 $$
$$ v_0- v_1=\dfrac{1}{3}d\omega_2 $$
Hence,
$$ \dfrac{ 3(v_0- v_1)}{d}= \omega_2 $$
Plugging into (1);
$$ v_0^2= v_1^2+ \frac{1}{6} d^2\left[ \dfrac{ 3(v_0- v_1)}{d} \right]^2 $$
$$ v_0^2= v_1^2+ \frac{1}{6} \color{red}{\bf\not}d^2 \dfrac{ 9(v_0- v_1)^2}{ \color{red}{\bf\not}d^2} $$
$$ v_0^2= v_1^2+ \dfrac{ 3(v_0- v_1)^2}{2 } $$
$$ 2 v_0^2=2 v_1^2+ 3(v_0^2-2v_0v_1 +v_1^2) $$
$$ 2 v_0^2=2 v_1^2+ 3 v_0^2-6v_0v_1 +3v_1^2 $$
$$ 5v_1^2-6v_0v_1+v_0^2 =0 $$
So that
$$v_1=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
where $a=5$, $b=-6v_0$, $c=v_0^2$
$$v_1=\dfrac{-(-6v_0)\pm\sqrt{(-6v_0)^2-(4\times 5\times v_0^2)}}{2\times 5}$$
$$v_1=\dfrac{ 6v_0 \pm\sqrt{36v_0^2-20 v_0^2}}{10}=\dfrac{6v_0 \pm 4v_0}{10}$$
whether $v_1=v_0$ which is rejected since it means that the rod will have no energy at all (the cube did not hit the rod at all).
Or $$\boxed{v_1=\frac{1}{5}v_0}$$
Notice that both velocities are to the right.
