Poisson's ratio is defined as 
$\mathrm{\sigma }=\frac{\mathrm{\Delta d}/\mathrm{d}}{\mathrm{\Delta \ell }/\mathrm{\ell }}\text{ or }\frac{\mathrm{\Delta \ell }}{\mathrm{\ell }}=\frac{\mathrm{\Delta d}/\mathrm{d}}{\mathrm{\sigma }}$ 
where d is the diameter and ℓ is the length of the wire. The area of cross-section is 
$\mathrm{A}={\mathrm{\pi r}}^2=\frac{{\mathrm{\pi d}}^2}{4}$ 
$\text{ or }\mathrm{logA}=\log \left(\frac{\mathrm{\pi }}{4}\right)+2\mathrm{logd}$ 
Differentiating, we have 
$\frac{\mathrm{\Delta A}}{\mathrm{A}}=2\frac{\mathrm{\Delta D}}{\mathrm{D}}$ 
 or    $\frac{\mathrm{\Delta d}}{\mathrm{d}}=\frac{1}{2}\frac{\mathrm{\Delta A}}{\mathrm{A}}=\frac{1}{2}\times 2\%=1\%$ 

 $(\because \frac{\mathrm{\Delta A}}{\mathrm{A}}=2\mathrm{\%},\mathrm{given})$ 
   Using this in (i) we have 
$\frac{\mathrm{\Delta \ell }}{\mathrm{\ell }}=\frac{1\%}{0.4}=2.5\%$