#### Answer

$\theta = 14.3^{\circ}$

#### Work Step by Step

The vertical component of the tension $T_y$ is equal in magnitude to the ball's weight. We can find $T_y$:
$T_y = mg$
$T_y = (0.0020~kg)(9.80~m/s^2)$
$T_y = 0.0196~N$
The horizontal component of tension $T_x$ is equal in magnitude to the electric force on the ball. We can find $T_x$:
$T_x = E~q$
$T_x = (200,000~N/C)(25\times 10^{-9}~C)$
$T_x = 0.0050~N$
We can find the angle $\theta$:
$tan(\theta) = \frac{T_x}{T_y}$
$\theta = arctan(\frac{T_x}{T_y})$
$\theta = arctan(\frac{0.0050~N}{0.0196~N})$
$\theta = 14.3^{\circ}$