A thin metallic wire having cross sectional area of 10 ^ - 4 m ^ 2...

The linear charge density of the ring is given by

$\lambda = \dfrac{Q}{2\piR} . . . \left(1\right)$

With reference to the above diagram, for the equilibrium condition of the ring, it can be written that

$2Tsin\left(\dfrac{d\theta}{2}\right) = \dfrac{1}{4\pi{ε}_{0}}\dfrac{{q}_{0}}{{R}^{2}} \cdot \left[\lambda\left(Rd\theta\right)\right]$

$⇒ T = \dfrac{1}{4\pi{ε}_{0}}\dfrac{{q}_{0}Q}{\left({R}^{2}\right) \times 2\pi}$ [As for very small angles $\sin \left(\dfrac{d\theta}{2}\right) = \dfrac{d\theta}{2}$]

$= \dfrac{\left(9 \times {10}^{9}\right)\left(2\pi \times 30 \times {10}^{- 12}\right)}{( 0 . 30{)}^{2} \times 2\pi}$

$= \dfrac{9 \times {10}^{- 3} \times 30}{9 \times {10}^{- 2}} = 3N$