this an easy one

Assume that the mass of the ball is m and the distance between the two balls is 2r (for easy calculation) and force between the balls be f

consider one of the balls(say the left one)

for this the weight is acting in downward direction and the electrostatic force due the other ball is in the horizontal direction.

given that the system is in equlibrium

the components of both the forces along the string are balanced by  tension(which we are no way conserned with)

now coming to the components perpendicular to the string they should inter compensate each other.

hence......

mg sin@ = f cos @

mg sin@ = [q^2/(4pie epsilon not)(2r)^2]cos@            --------------(say eq 1)

but here sin@ =(2r/2)/l ( by considering the triangle formed by pt of suspension the ball and the mid point of the balls)

hence r= l sin@                                                                   --------------(say eq 2)

substituting value of eq 2 in eq 1, we have

mg sin@ = [q^2/(4 pie epsilon not)(4(l sin@)^2)]cos@

mg sin@ = [q^2/(16)(pie epsilon not)(l^2)(sin^2@)]cos@

taking  g sin@ the other side

m =  [q^2/(g)(16)(pie epsilon not)(l^2)(sin^2@)]*(cos@/sin@)

m =  [q^2/(g)(16)(pie epsilon not)(l^2)(sin^2@)]cot@

(sorry wasnt able to give the diagram due to some error, hope u understood the sum)