The elastic potential energy stored in a steel wire of length $$20$$ m stretched through $$2$$ cm is $$80$$ J. The cross sectional area of the wire is _____ mm$$^2$$. (Given, $$Y = 2.0 \times 10^{11}$$ N m$$^{-2}$$)

Given: Length $$L = 20$$ m, extension $$\Delta L = 2$$ cm $$= 0.02$$ m, elastic potential energy $$U = 80$$ J, and Young's modulus $$Y = 2.0 \times 10^{11}$$ N/m$$^2$$.

The elastic potential energy stored in a stretched wire is:

$$U = \frac{1}{2} \times \frac{YA(\Delta L)^2}{L}$$

Solving for the cross-sectional area $$A$$:

$$A = \frac{2UL}{Y(\Delta L)^2}$$

Substituting the values:

$$A = \frac{2 \times 80 \times 20}{2.0 \times 10^{11} \times (0.02)^2}$$

$$= \frac{3200}{2.0 \times 10^{11} \times 4 \times 10^{-4}}$$

$$= \frac{3200}{8 \times 10^{7}} = 4 \times 10^{-5} \text{ m}^2$$

Converting to mm$$^2$$:

$$A = 4 \times 10^{-5} \text{ m}^2 = 4 \times 10^{-5} \times 10^6 \text{ mm}^2 = 40 \text{ mm}^2$$

Therefore, the answer is $$\mathbf{40}$$.