Answer
$q = +1.78\times 10^{-7}~C$
Work Step by Step
The vertical component of the tension in the string is equal in magnitude to the weight of the ball:
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
We can find the charge $q$:
$q~E = F_T~sin~\theta$
$q~E = \frac{mg~sin~\theta}{cos~\theta}$
$q = \frac{mg~tan~\theta}{E}$
$q = \frac{(0.0050~kg)(9.8~m/s^2)~tan~20^{\circ}}{10^5~N/C}$
$q = +1.78\times 10^{-7}~C$
