$\text{At distance of closest approach relative velocity of the two particles are equal} $

 $\text{ Let this velocity is equal to v'}.$

$\text{From momentum conservation } 2mv' = mv , v'= \frac{v}{2}$

$ \text{From Energy conservation } \frac{1}{2} mv^2 = 2\times \frac{1}{2} m(\frac{v}{2}^2) + \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r}$

$\frac{Q^2}{4\pi\epsilon_0r} = \frac{1}{2} mv^2 - \frac{1}{4} mv^2 = \frac{1}{4} mv^2$

$ r = \frac{1}{4\pi\epsilon_0} \frac{4Q^2}{mv^2}$