In vertical derection
2T sin = 20
using small angle approximation $\sin \theta=\theta$
$ \begin{aligned} & \theta=\frac{1}{100} \\ & \therefore \quad T=\frac{10}{\theta} \\ & \quad T=1000 N \end{aligned} $
$ \text { Change in length } \Delta \mathrm{L} \quad \begin{aligned} & =2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{~L} \\ & =2 \mathrm{~L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{~L}^2}-1\right] \\ \Delta \mathrm{L} & =\frac{\mathrm{x}^2}{\mathrm{~L}} \end{aligned} $
$ \begin{aligned} & \therefore \text { Modulus of elasticity }=\frac{\text { stress }}{\text { strain }} \\ & 2 \times 10^{11}=\frac{10^3}{\mathrm{~A} \times \frac{\mathrm{x}^2}{\mathrm{~L}}} \times 2 \mathrm{~L} \\ & \therefore A=1 \times 10^{-4} \mathrm{~m}^2 \end{aligned} $