$\because$ Stress, $\frac{F}{∆A}=1\%$ of $Y=\frac{Y}{100}$ But $\mathrm{Y}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{g}\mathrm{\prime }\mathrm{s}$ modulus, $Y=\frac{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}}{\mathrm{s}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}}=\frac{\frac{F}{∆A}}{\frac{∆l}{l}}$

$\therefore Y=\frac{\frac{Y}{100}}{\frac{∆l}{l}} \left(\mathrm{p}\mathrm{u}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{n}\mathrm{g}\frac{F}{∆A}=\frac{Y}{100}\right)$

$\therefore \frac{∆l}{l}=\frac{1}{100}$

${\mathrm{P}\mathrm{o}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{n}}^{\mathrm{\prime }}\mathrm{s}$ ratio, $\sigma =\frac{-\frac{∆r}{r}}{\frac{∆l}{l}}$

$\therefore \frac{∆r}{r}=-\sigma \cdot \frac{∆l}{l}=\frac{-0.3}{100}$

$\frac{∆r}{r}=\frac{-0.3}{100}$

$\therefore$ Change in volume, $\frac{∆V}{V}=\frac{2∆r}{r}+\frac{∆l}{l}$

$=\frac{2\times \left(-0.3\right)}{100}+\frac{1}{100}$

$=\frac{1-0.6}{100}=\frac{0.4}{100}$

So the percentage change in volume = $0.4\%$