Suppose a uniformly charged wall provides a uniform electric field...

As we can see from geometry, $sin\theta = \dfrac{10}{20} = \dfrac{1}{2}$

$⇒ \theta = 30 °$

Vertical component of tension will balance the weight and the horizontal component will balance the electric force. Therefore,

$tan\theta = \dfrac{qE}{mg}$

$⇒ \tan 30 ° = \dfrac{q \times 2 \times {10}^{4}}{2 \times {10}^{- 3} \times 10}$

$⇒ \dfrac{1}{\sqrt{3}} = q \times {10}^{6}$

$⇒ q = \dfrac{1}{\sqrt{3}} \times {10}^{- 6}C = \dfrac{1}{\sqrt{3}}\mu C$

Hence, $x = 3$.