Answer
$T = 0.40~N$
Work Step by Step
We can find the angle $\theta$ the string makes with the vertical:
$sin~\theta = \frac{r}{L}$
$sin~\theta = \frac{C/(2\pi)}{L}$
$sin~\theta = \frac{C}{2\pi~L}$
$sin~\theta = \frac{0.94~m}{(2\pi)(0.90~m)}$
$\theta = sin^{-1}~\frac{0.94~m}{(2\pi)(0.90~m)}$
$\theta = 9.57^{\circ}$
The vertical component of the tension is equal and opposite to the weight of the bob. We can find the tension $T$ in the string:
$T_y = mg$
$T~cos~\theta = mg$
$T = \frac{mg}{cos~\theta}$
$T = \frac{(0.040~kg)(9.8~m/s^2)}{cos~9.57^{\circ}}$
$T = 0.40~N$
