There is no external force on system (ball A and ball B). The balls moves due to mutual attraction that is internal force to the system. The centre of mass of the system will not move. The will collide at the position of centre of mass of balls A and B. It is easy to calculate the position of centre of mass at initial stage.
${\mathrm{x}}_{\mathrm{cm}} = \frac{{\mathrm{m}}_{\mathrm{A}}\times \mathrm{A}+{\mathrm{m}}_{\mathrm{B}}\times \mathrm{B}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}} = \frac{{\mathrm{m}}_{\mathrm{A}}.\mathrm{o}+{\mathrm{m}}_{\mathrm{B}}.10\mathrm{R}}{({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}})}$

$m_A = \frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_{\mathrm{A}} \mathrm{and} {\mathrm{m}}_{\mathrm{B}} = \frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_{\mathrm{B}}$

$\mathrm{As} {\mathrm{\rho }}_{\mathrm{A}} =2{\mathrm{\rho }}_{\mathrm{B}} \Rightarrow {\mathrm{m}}_{\mathrm{A}} = 2{\mathrm{m}}_{\mathrm{B}}$

Hence ${\mathrm{x}}_{\mathrm{cm}} = \frac{{\mathrm{m}}_{\mathrm{B}}\left(10\mathrm{R}\right)}{3{\mathrm{m}}_{\mathrm{B}}} = \frac{10}{3}\mathrm{R}$

When they are touching each other then also the centre of mass lie at same place.

X'_cm = $\frac{{\mathrm{m}}_{\mathrm{A}}\left(0\right)+{\mathrm{m}}_{\mathrm{B}}2\mathrm{R}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}=\frac{2\mathrm{R}}{3}$ from the first

The two are touching each other at an additional $\frac{R}{3}$ distance from CM.

Hence the collision is happening at $\frac{10R}{3}+\frac{R}{3}=\frac{11R}{3}$