If $\alpha$ and $\beta$ be the longitudinal strain and lateral strain respectively, then
$\alpha=\frac{l_{2}-l_{1}}{l_{1}}=10^{-3}=0.001$
$\Rightarrow \quad \frac{l_{2}}{l_{1}}=1.001$
Poisson's ratio, $\sigma=\frac{\text { Lateral strain }}{\text { Longitudinal strain }}$
$=\frac{\beta}{\alpha}$
$\begin{array}{ll}\Rightarrow & \beta=\sigma \alpha=0 \\ \text { But, } & \beta=\frac{r_{1}-r_{2}}{r_{1}}\end{array}$
$\therefore \quad \frac{r_{1}-r_{2}}{n_{1}}=10^{-4}=0.0001$
$\Rightarrow \quad \frac{r_{2}}{r_{1}}=0.9999$
Hence, percentage increase in vol
$=\frac{V_{2}-V_{1}}{V_{1}} \times 100$
$=\frac{\pi r_{2}^{2} l_{2}-\pi r_{1}^{2} l_{1}}{\pi r_{1}^{2} l_{1}} \times 100$
$=\left[\left(\frac{r_{2}}{r_{1}}\right)^{2}\left(\frac{l_{2}}{l_{1}}\right)^{2}-1\right] \times 100$
$=\left[(0.9999)^{2} \times(1.0001)^{2}-1\right] \times 100$
$=0.0008 \times 100=0.08 \%$