First, we need to find the maximum force that can be applied to the steel wire within its elastic limit. This force can be calculated using the given area under stress and the stress limit provided by the elastic limit of steel.

Let's assume the area of cross-section of the wire is $A$. The force exerted can be given by:

$ F = \sigma \times A $

where:

$$ \sigma = 8 \times 10^8 \mathrm{~N/m}^2 $$ (Elastic limit of steel)

To calculate the elongation ($ \Delta L $) under this force, we use Hooke's Law, which relates force, elongation, cross-sectional area, original length, and Young's modulus as follows:

$$ F = \frac{Y \times A \times \Delta L}{L} $$

where:

$$ Y = 2 \times 10^{11} \mathrm{~N/m}^2 $$ (Young's modulus of steel),

$ L = 1 \mathrm{~m} $ (original length of the wire).

Substituting for $ F $ from the earlier expression and rearranging the formula, we get:

$$ \sigma \times A = \frac{Y \times A \times \Delta L}{L} $$

$$ \sigma = \frac{Y \times \Delta L}{L} $$

$$ \Delta L = \frac{\sigma \times L}{Y} $$

$$ \Delta L = \frac{8 \times 10^8 \times 1}{2 \times 10^{11}} \mathrm{~meters} $$

$$ \Delta L = 0.004 \mathrm{~meters} $$

$$ \Delta L = 4 \mathrm{~mm} $$

Thus, the maximum elongation of the wire within the elastic limit is $4 \mathrm{~mm}$.

The answer is: Option A - 4 mm.