To solve this problem, we need to determine the maximum elongation of a steel wire under its elastic limit.$\\$Given:$\\$• Length of the steel wire $L = 1$ m$\\$• Elastic limit of steel $\sigma = 8 \times 10^8$ N/m$^2 \\$• Young's modulus of steel $Y = 2 \times 10^{11}$ N/m$^2 \\$The formula for elongation $\Delta L$ under stress is given by:$\\ \Delta L = \frac{\sigma \cdot L}{Y} \\$Substitute the given values:$\\ \Delta L = \frac{8 \times 10^8 \cdot 1}{2 \times 10^{11}} \\$Simplify the expression:$\\ \Delta L = \frac{8}{2} \times 10^{-3} \\ \Delta L = 4 \times 10^{-3}$ m$\\$Convert the elongation from meters to millimeters:$\\ \Delta L = 4 \times 10^{-3} \times 1000$ mm/m$\\ \Delta L = 4$ mm$\\$Therefore, the maximum elongation of the steel wire is $4$ mm.$\\$This corresponds to Option 3.