A vertical glass tube 1.20 m tall, with an inner radius of 2.5 cm, is full of water to the brim…

Step 1. Wavelength (a). $\lambda = \dfrac{c}{f} = \dfrac{336\ \text{m/s}}{512\ \text{Hz}} = 65.6\ \text{cm}$.

Step 2. The first resonance (b). The end correction is $0.6 \times 2.5\ \text{cm} = 1.5\ \text{cm}$. A closed pipe first resonates when $l + e = \lambda/4$:

Step 3. Its time (c). The water must fall $14.91\ \text{cm}$ at $2.0\ \text{cm/s}$: $t_1 = 7.45\ \text{s}$.

Step 4. The next one (d). The next resonance is at $l + e = 3\lambda/4$, half a wavelength further down, $32.81\ \text{cm}$. That takes $\dfrac{32.81\ \text{cm}}{2.0\ \text{cm/s}} = 16.4\ \text{s}$.

Step 5. How many (e). The resonance lengths are $l = (2n - 1)\lambda/4 - e$. The longest one below $120\ \text{cm}$ is $113.3\ \text{cm}$, for $n = 4$; the next would be $146.2\ \text{cm}$, longer than the tube. So $4$ resonances are heard.