The correct answer is 2.30.

Step 1: Initial Parameters

Let the initial radius of stars A and B be $R$.
Initial mass of star A is $M_A=\frac{4}{3}\pi R^3{\rho }_A$
Initial mass of star B is $M_B=2M_A$

Step 2: Properties of Star A After Interaction

The radius of star A becomes $R_A^{\prime }=\frac{R}{2}$ while keeping density ${\rho }_A$ constant.
New mass of star A:

$M_A^{\prime }=\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3{\rho }_A=\frac{M_A}{8}$

Mass lost by star A:

$\Delta M=M_A-\frac{M_A}{8}=\frac{7}{8}{M}_A$

Escape velocity from star A after interaction:

$v_A=\sqrt{\frac{2G{M}_A^{\prime }}{R_A^{\prime }}}=\sqrt{\frac{2G\left(\frac{M_A}{8}\right)}{\frac{R}{2}}}=\sqrt{\frac{G{M}_A}{2R}}$

Step 3: Properties of Star B After Interaction

New mass of star B:

$M_B^{\prime }=M_B+\Delta M=2M_A+\frac{7}{8}$

The added shell has density ${\rho }_A$, so the outer radius $R_B^{\prime }$ of B is given by:

$\frac{4}{3}\pi {\left(R_B^{\prime }\right)}^3-\frac{4}{3}\pi R^3=\frac{\Delta M}{{\rho }_A}=\frac{7}{8}\left(\frac{4}{3}\pi R^3\right)$

${\left(R_B^{\prime }\right)}^3=R^3+\frac{7}{8}{R}^3=\frac{15}{8}{R}^3\Rightarrow R_B^{\prime }=\frac{{15}^{\frac{1}{3}}}{2}R$

Escape velocity from star B after interaction:

$v_B=\sqrt{\frac{2G{M}_B^{\prime }}{R_B^{\prime }}}=\sqrt{\frac{2G\left(\frac{23M_A}{8}\right)}{\frac{{15}^{\frac{1}{3}}}{2}R}}=\sqrt{\frac{23G{M}_A}{2·{15}^{\frac{1}{3}}R}}$

Step 4: Finding the Ratio and Solving for n

$\frac{v_B}{v_A}=\frac{\sqrt{\frac{23G{M}_A}{2·{15}^{\frac{1}{3}}R}}}{\sqrt{\frac{G{M}_A}{2R}}}=\sqrt{\frac{23}{{15}^{\frac{1}{3}}}}$

Equating this to the given relation:

$\sqrt{\frac{23}{{15}^{\frac{1}{3}}}}=\sqrt{\frac{10n}{{15}^{\frac{1}{3}}}}$

$10n=23\Rightarrow n=2.30$