Answer
$q = 7.46\times 10^{-7}~C$
Work Step by Step
The vertical component of the tension in each thread is equal in magnitude to the weight of one sphere:
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
We can find the charge $q$:
$\frac{kq^2}{r^2} = F_T~sin~\theta$
$\frac{kq^2}{(2~L~sin~\theta)^2} = (\frac{mg}{cos~\theta})~(sin~\theta)$
$q^2 = \frac{4L^2~mg~tan~\theta~sin^2~\theta}{k}$
$q = 2L~sin~\theta~\sqrt{\frac{mg~tan~\theta}{k}}$
$q = (2)(1.0~m)~sin~20^{\circ}~\sqrt{\frac{(0.0030~kg)(9.8~m/s^2)~tan~20^{\circ}}{9.0\times 10^9~N~m^2/C^2}}$
$q = 7.46\times 10^{-7}~C$
