There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is:

Volume of cylindrical wire, V = $\frac{{\mathrm{\pi x}}^2\mathrm{L}}{4}$, where x is diameter of wire

Differentiating both sides

$\frac{\mathrm{dV}}{\mathrm{dx}} = \frac{\mathrm{\pi }}{4}[2\mathrm{xL}+{\mathrm{x}}^2.\frac{\mathrm{dL}}{\mathrm{dx}}]$

Also volume remains constant    $\frac{\mathrm{dV}}{\mathrm{dx}} = 0$

2xL +${\mathrm{x}}^2\frac{\mathrm{dL}}{\mathrm{dx}}= 0 \Rightarrow 2\mathrm{xL} = -{\mathrm{x}}^2\frac{\mathrm{dL}}{\mathrm{dx}}$

$\Rightarrow$ $\frac{{\displaystyle \frac{\mathrm{dx}}{\mathrm{x}}}}{{\displaystyle \frac{\mathrm{dL}}{\mathrm{L}}}} = -\frac{1}{2}$

$\Rightarrow$Poisson's ratio = -$\frac{1}{2}$

So the correct choice is (b).