The force required to stretch a wire of cross-section 1 cm$$^2$$ to double its length will be: (Given Young's modulus of the wire $$= 2 \times 10^{11}$$ N m$$^{-2}$$)

To determine the force required to stretch a wire to double its length, we note that the cross-sectional area is $$A = 1 \text{ cm}^2 = 1 \times 10^{-4} \text{ m}^2$$ and Young's modulus is $$Y = 2 \times 10^{11}$$ N/m$$^2$$. Since the extension is equal to the original length, $$\Delta L = L$$, the strain is $$\frac{\Delta L}{L} = 1$$.

Young's modulus is given by the relation $$Y = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L}$$. Substituting $$\Delta L = L$$ yields $$F = Y \times A \times \frac{\Delta L}{L}$$.

Substituting the numerical values, we get $$F = 2 \times 10^{11} \times 1 \times 10^{-4} \times 1 = 2 \times 10^{7} \text{ N}$$. Hence, the correct answer is Option C: $$2 \times 10^7$$ N.