#### Answer

(a) $F_{elec} = 0.0036~N$
(b) $T = 0.0104~N$

#### Work Step by Step

The vertical component of the tension $T_y$ is equal in magnitude to the ball's weight. We can find the horizontal component of the tension $T_x$. So;
$\frac{T_x}{T_y} = tan(\theta)$
$T_x = T_y~tan(\theta)$
$T_x = mg~tan(\theta)$
$T_x = (0.0010~kg)(9.80~m/s^2)~tan(20^{\circ})$
$T_x = 0.0036~N$
The force $F_{elec}$ is equal in magnitude to the horizontal component of the tension $T_x$. So;
$F_{elec} = T_x$
$F_{elec} = 0.0036~N$
(b) We can find the tension $T$ in the string.
$cos(\theta) = \frac{T_y}{T}$
$T = \frac{T_y}{cos(\theta)}$
$T = \frac{mg}{cos(\theta)}$
$T = \frac{(0.0010~kg)(9.80~m/s^2)}{cos(20^{\circ})}$
$T = 0.0104~N$