A copper wire 2.5 m long and 1.6 mm across is stretched by 1.2 mm…

Step 1. $\varepsilon = \dfrac{1.2 \times 10^{-3}\ \text{m}}{2.5\ \text{m}} = 4.8 \times 10^{-4}$.

Step 2. Lateral strain $= 0.34 \times 4.8 \times 10^{-4} = 1.63 \times 10^{-4}$, so $\Delta d = 1.63 \times 10^{-4} \times 1.6\ \text{mm} = 2.61 \times 10^{-4}\ \text{mm} = 0.261\ \mu\text{m}$.

Step 3. Volume. $V = \pi(0.80 \times 10^{-3}\ \text{m})^2 \times 2.5\ \text{m} = 5.03 \times 10^{-6}\ \text{m}^3$, and $\dfrac{\Delta V}{V} = 4.8 \times 10^{-4} \times 0.32 = 1.54 \times 10^{-4}$, so $\Delta V = 7.72 \times 10^{-10}\ \text{m}^3$.