Let the initial speed of $m_1$ is $v_0$, then from the question we know that

$\frac{4}{9}\frac{1}{2}{m}_1{v_0}^2=\frac{1}{2}{m}_1{v_1}^2$

$v_1=\frac{2}{3}{v}_0$

Since the collision is elastic, $e=1$.

$\Rightarrow v_0=v_2+v_1$

where $v_1$ is the speed of $m_1$ after the collision and $v_2$ is the speed of $m_2$ after collision.

$\Rightarrow v_2=\frac{v_0}{3}$

$m_1{v}_0=-m_1\frac{2}{3}{v}_0+m_2\frac{v_0}{3}$

$\frac{m_1}{m_2}=\frac{1}{5}$