Given, energy per unit volume = $$\frac{1}{2} \times \text{stress} \times \text{strain}$$

The stress can be given as $$\text{stress} = Y \times \text{strain}$$, where Y is the Young's modulus.

The energy stored in the wire can be written as:

$$\text{Energy} = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$$

Substituting the stress formula, we get:

$$\text{Energy} = \frac{1}{2} \times Y \times \text{strain}^2 \times A \times L$$

We are given that the energy stored is $$80 \ \text{J}$$, the original length of the wire is $$20 \ \text{m}$$, the elongation is $$2 \ \text{cm}$$, and the Young's modulus is $$2.0 \times 10^{11} \ \text{Nm}^{-2}$$. We need to find the cross-sectional area (A) of the wire.

$$80 = \frac{1}{2} \times 2 \times 10^{11} \times \left(\frac{2 \times 10^{-2}}{20}\right)^2 \times A \times 20$$

Now we can solve for A:

$$A = \frac{80 \times 20^2}{(2.0 \times 10^{11}) \times (2 \times 10^{-2})^2} = 40 \times 10^{-6} \ \text{m}^2$$

To convert the area to $$\text{mm}^2$$, we multiply by $$10^6$$:

$$A = 40 \times 10^{-6} \times 10^6 = 40 \ \text{mm}^2$$

So, the cross-sectional area of the wire is $$40 \ \text{mm}^2$$.