If $A_1$ and $A_2$ are the mass number of two parts.
The radius of nucleus is given by
$R=R_0(A)^{1 / 3}$
So, $R_1=R_0\left(A_1\right)^{1 / 3}$ and $R_2=R_0\left(A_2\right)^{1 / 3}$
$ \Rightarrow \frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3} $
$ \Rightarrow \frac{A_1}{A_2}=\left(\frac{R_1}{R_2}\right)^3=\left(\frac{1}{2}\right)^3=\frac{1}{8} $
$\therefore \text { Ratio of masses, } \frac{m_1}{m_2}=\frac{A_1}{A_2}=\frac{1}{8}$
From conservation of momentum,
$ \Rightarrow m_1 v_1=m_2 v_2$
$\Rightarrow \frac{v_1}{v_2}=\frac{m_2}{m_1}=\frac{8}{1} \text { or } 8: 1$