#### Answer

$q = +1.8\times 10^{-7}~C$

#### Work Step by Step

The vertical component of the tension $T_y$ is equal in magnitude to the ball's weight. We can find $T_y$:
$T_y = mg$
$T_y = (0.0050~kg)(9.80~m/s^2)$
$T_y = 0.049~N$
We can find the horizontal component of the tension $T_x$:
$\frac{T_x}{T_y} = tan(\theta)$
$T_x = T_y~tan(\theta)$
$T_x = (0.049~N)~tan(20^{\circ})$
$T_x = 0.018~N$
The horizontal component of tension is equal in magnitude to the electric force on the ball. We can find the magnitude of the charge on the ball.
$E~q = T_x$
$q = \frac{T_x}{E}$
$q = \frac{0.018~N}{100,000~N/C}$
$q = 1.8\times 10^{-7}~C$
Since the electric force on the ball is directed to the right, the charge on the ball is positive.
$q = +1.8\times 10^{-7}~C$