A piston is inserted into a cylindrical tube of small cross section...

The distance between two consecutive resonant lengths in a closed organ pipe is equal to half the wavelength of the sound wave.

$\Delta L = \dfrac{\lambda}{2}$

Given that the displacement of the piston between the first and second resonance is $32.0 \text{ cm}$, we have:

$\Delta L = 32.0 \text{ cm} = 0.32 \text{ m}$

$\dfrac{\lambda}{2} = 0.32 \Rightarrow \lambda = 0.64 \text{ m}$

The speed of sound $v$ is given by the relation:

$v = f \lambda$

Substituting the given frequency $f = 512 \text{ Hz}$ and the calculated wavelength $\lambda = 0.64 \text{ m}$:

$v = 512 \times 0.64 = 327.68 \text{ m/s}$

Rounding off to the nearest integer, we get $v \approx 328 \text{ m/s}$.