Let us consider the diagram as shown above. The forces acting on the system are:

• Gravitational force : The mass is indicated by green ball and because of the weight there is a gravitational force acting in the downward direction.
• Coulomb’s force : Since the two charges are equal in magnitude, there will be force of repulsion as given by couloumb law
• The tension acting on the string T can be resolved into horizontal and vertical components

We can draw a freefall diagram depicting the entire picture with the above 3 forces. Now we have to balance the horizontal forces and vertical forces separately.

Vertical forces are (a) Force mg acting in the downward direction (b) The vertical component of tension T acting on the string which is Tsinθ

Hence,  Tsinθ = mg --------Eqn (1)

Horizontal forces are (a) Force Fr which is coulomb force due to repulsion (b) The horizontal  component of tension T acting on the string which is Tcosθ

Hence, Tcosθ = F_r = kq₁q₂ / (2r)² --------Eqn (2)

Where 2r is the distance between the balls

Now, from triangle ACD, sin θ = r/L Hence, r² = (Lsinθ)²

As the charges are equal, we can consider q_{1 =}  q₂ = q

Hence, Eqn (2) becomes Tcosθ = kq² / 4L²sin²θ  ------------Eqn (3)

Eqn (1) / Eqn (2) gives Tsinθ / Tcosθ  = mg * 1/ (  kq² / 4L² sin²θ)

q²  = (mg /k tan θ) * (4L ²sin²θ)

As k = 8.99 * 10⁹ Nm² / couloumb

Also,  given m = 0.2 * 10^-3 kg, L = 50 * 10^-2m, g = 9.8 m/s² and θ = 37 ⁰

q² = 0.2 * 10^-3 * 9.8 * 4 * 50 * 10^-2 * 50 * 10^-2 * sin²37 / (8.99 * 10⁹) tan 37

q² = 937.49 * 10^-16

Finding the square root of the above expression, gives the value of q

q = 3.062 * 10^-7 C