1. Given data

• Young’s modulus: $Y = 2\times 10^{11}\,\text{N m}^{-2}$
• Original length of wire: $2\,\text{m}$
• Mass suspended at middle: $m=2\,\text{kg}$
• Hence weight: $mg = 20\,\text{N}$
• Angle made at each support: $\theta = \dfrac{1}{100}\,\text{rad}$
• Required: cross-sectional area $A$

Since the mass is suspended at the middle, the wire is divided into two equal halves. So original length of each half is $L=1\,\text{m}.$

2. Tension in each half of wire

At the mass, vertical equilibrium gives $2T\sin\theta = mg.$

For small angle, $\sin\theta \approx \theta = \frac{1}{100}$. So $2T\left(\frac{1}{100}\right)=20$ $\frac{2T}{100}=20$ $T=1000\,\text{N}.$

3. Extension of each half due to tension

Extension formula: $\Delta L = \frac{TL}{AY}.$

Here $L=1\,\text{m}$, so

4. Geometrical relation for small sag

If the midpoint sags and each half makes angle $\theta$ with the horizontal, then new length of each half is $L' = \frac{L}{\cos\theta}.$

Hence extension in each half is $\Delta L = L'-L = L\left(\frac{1}{\cos\theta}-1\right).$

For small $\theta$, $\frac{1}{\cos\theta}-1 \approx \frac{\theta^2}{2}.$

Thus $\Delta L \approx L\frac{\theta^2}{2} = 1\cdot \frac{(1/100)^2}{2} = \frac{1}{2\times 10^4}=5\times 10^{-5}\,\text{m}.$

5. Equate elastic extension and geometrical extension

$\frac{1000}{A\cdot 2\times 10^{11}} = 5\times 10^{-5}.$

So, $1000 = A\cdot 2\times 10^{11}\cdot 5\times 10^{-5}.$

Now, $2\times 10^{11}\cdot 5\times 10^{-5} = 10\times 10^6 = 10^7.$

Therefore, $1000 = A\times 10^7$ $A = \frac{1000}{10^7} = 10^{-4}\,\text{m}^2.$

6. Final value

The question asks for $A$ in the form $\underline{\hspace{1cm}}\times 10^{-4}\,\text{m}^2.$

So the required integer is $\boxed{1}.$