One end of a silk thread of length 10 cm is attached to the large...

In equilibrium, the forces acting on the small ball are the weight of the ball $mg$ acting vertically downwards, the electric force $qE$ acting horizontally, and the tension $T$ in the silk thread acting along the thread.

Resolving the tension $T$ into vertical and horizontal components gives $T \cos\theta$ and $T \sin\theta$ respectively.

For vertical equilibrium, the vertical component of tension balances the weight of the ball:

$T \cos\theta = mg$

Substituting the given values $m = 10 \text{ g} = 0.01 \text{ kg}$, $g = 10 \text{ m/s}^2$, and \theta = 60^{}^{{}^{\circ}}:

T \cos 60^{}^{{}^{\circ}} = 0.01 \times 10

$T \times \dfrac{1}{2} = 0.1$

$T = 0.20 \text{ N}$