Two spherical stars A and B have densities ρ _ A and ρ _ B ,...

Given here: ${R}_{A} = {R}_{B} = R$ and ${M}_{B} = 2{M}_{A}$.

Now, after interaction process, radius of remaining star $A$ is ${{R}_{A}}^{'} = \dfrac{R}{2}$ and its mass is ${{M}_{A}}^{'} = {ρ}_{A}\dfrac{4}{3}\pi{\left(\dfrac{{R}_{A}}{2}\right)}^{3} = \dfrac{{M}_{A}}{8}$

Applying conservation of energy,

$\dfrac{- G{{M}_{A}}^{'}m}{{{R}_{A}}^{'}} + \dfrac{1}{2}m{v}_{A}^{2} = 0$

Escape velocity of star $A$ is ${v}_{A} = \sqrt{\dfrac{2G{M}_{A}}{8 \times \left(\dfrac{R}{2}\right)}} = \dfrac{{v}_{0}}{2}$

Now, for $B,$ mass collected over $B$ is ${{M}_{B}}^{'} = {M}_{A} - \dfrac{{M}_{A}}{8} = \dfrac{7}{8}{M}_{A}$.

Let the radius of star $B$ after interaction becomes $r$.

Applying mass conservation,

$\dfrac{4}{3}\pi\left({r}^{3} - {R}^{3}\right){ρ}_{A} = \dfrac{4}{3}\pi{R}^{3} \times \dfrac{7}{8}{ρ}_{A}$

$⇒ r = \left(\dfrac{15}{8}\right)\dfrac{1}{3}R$

Escape velocity of star $B$ is

$∴ {v}_{B} = \sqrt{\dfrac{2G \times \left(2{M}_{A} + \dfrac{7}{8}{M}_{A}\right)}{{\left(15\right)}^{1 / 3}\left(\dfrac{R}{2}\right)}}$

$= \sqrt{\dfrac{2G{M}_{A}}{R}}\sqrt{\dfrac{2\left(\dfrac{16 + 7}{8}\right)}{{\left(15\right)}^{\dfrac{1}{3}}}}$

$= {v}_{0} \times \sqrt{\dfrac{23 \times 2}{8 \times \left(15\right)\dfrac{1}{3}}} = \dfrac{{v}_{0}}{2} \times \sqrt{\dfrac{23}{\left(15\right)\dfrac{1}{3}}}$

Now, the ratio $\dfrac{{v}_{B}}{{v}_{A}} = \sqrt{\dfrac{23}{\left(15\right)\dfrac{1}{3}}} = \sqrt{\dfrac{2 . 30 \times 10}{\left(15\right)\dfrac{1}{3}}}$

$∴ n = 2 . 30$