The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have

$${m_1}{v_1} = {m_2}{v_2} \Rightarrow {{{v_1}} \over {{v_2}}} = {{{m_2}} \over {{m_1}}} \Rightarrow {{{m_2}} \over {{m_1}}} = {8 \over {27}}$$

Now, since $$m = \rho {4 \over 3}\pi {r^3}$$

Therefore, $${{{m_2}} \over {{m_1}}} = {{\rho {4 \over 3}\pi r_2^3} \over {\rho {4 \over 3}\pi r_1^3}} \Rightarrow {{{m_2}} \over {{m_1}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} \Rightarrow {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} = {8 \over {27}}$$

$$ \Rightarrow {{{r_2}} \over {{r_1}}} = {2 \over 3}$$

Thus, ratio of radii of nuclei $${r_1}:{r_2} = 3:2$$.