Given here, $\frac{v_1}{v_2}=\frac{3}{2}$

Using conservation of momentum during disintegration, we have

$m_1{v}_1=m_2{v}_2\Rightarrow \frac{m_1}{ m_2}=\frac{2}{3}$

Since, $m=\frac{4}{3}\pi R^3\rho$

We have $m\propto R^3 \left(\because \text{nuclear mass density}, \mathrm{\rho }=\text{constant}\right)$

$\frac{m_1}{m_2}={\left(\frac{R_1}{R_2}\right)}^3$

$\Rightarrow \frac{R_1}{R_2}={\left(\frac{2}{3}\right)}^{\frac{1}{3}}$

So, $x=2$.