To find the elastic potential energy density of the steel wire, we need to use the given information and formulae for strain and energy density.

Given:

The length of the wire, $ \ell = 2 \, \text{m} $

Young's modulus, $ Y = 2.0 \times 10^{11} \, \text{N/m}^2 $

Poisson's ratio, $ \mu = 0.2 $

Transverse strain, $ \frac{\Delta r}{r} = 10^{-3} $

The formula for Poisson's ratio is:

$ \mu = -\frac{\left(\frac{\Delta r}{r}\right)}{\left(\frac{\Delta \ell}{\ell}\right)} $

From this, we solve for the longitudinal strain $\frac{\Delta \ell}{\ell}$:

$ \frac{\Delta \ell}{\ell} = \frac{1}{\mu} \times \left(\frac{\Delta r}{r}\right) $

Substitute the given values:

$ \frac{\Delta \ell}{\ell} = \frac{1}{0.2} \times 10^{-3} = 5 \times 10^{-3} $

The elastic potential energy density $ u $ is given by:

$ u = \frac{1}{2} Y \varepsilon_{\ell}^2 $

where $ \varepsilon_{\ell} = \frac{\Delta \ell}{\ell} $. Plug in the values:

$ u = \frac{1}{2} \times 2 \times 10^{11} \times \left(5 \times 10^{-3}\right)^2 $

Simplify further:

$ u = \frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{-6} $

$ u = 25 \times 10^5 \, \text{(in SI units)} $

Thus, the elastic potential energy density of the wire is $ 25 \times 10^5 $ SI units.