B. 3.83 cm\(^3\)

The bore is a cavity, so its volume increases with temperature, like the material's volume expansion.

Volume expansion is given by: \(\Delta V = V_0 \gamma \Delta T\)

where \( \gamma = 3\alpha \) and \( \alpha = 24 \times 10^{-6} \, \text{K}^{-1} \), so:
\(\gamma = 72 \times 10^{-6} \, \text{K}^{-1}\)

The temperature change is: \(\Delta T = 340 - 34 = 306 \, \text{K}\)

Now, substituting for \( \Delta V \):
\(\Delta V = 3.48 \times (72 \times 10^{-6}) \times 306 \approx 0.352 \, \text{cm}^3\)

The new volume \( V \) is: \(V = 3.48 + 0.352 \approx 3.832 \, \text{cm}^3 \approx 3.83 \, \text{cm}^3\).