So using conservation of momentum we get

$m v=2 m u$ or $u=\frac{v}{2}$

The initial energy of the system is given as $\frac{1}{2} m v^{2}$

and the energy at the minimum distance is given as

$\frac{1}{2} m\left(\frac{v}{2}\right)^{2}+\frac{1}{2} m\left(\frac{v}{2}\right)^{2}+\frac{1}{4 \pi \in_{o}} \frac{Q^{2}}{R}$

Equating the two energies we get

$\frac{1}{2} m v^{2}=\frac{1}{4} m v^{2}+\frac{1}{4 \pi \in_{o}} \frac{Q^{2}}{R}$

Or

$\frac{1}{4} m v^{2}=\frac{1}{4 \pi \in_{o}} \frac{Q^{2}}{R}$

Or

$R=\frac{4 Q^{2}}{4 \pi \in_{o} m v^{2}}$