Answer
635 Hz.
Work Step by Step
The column of air is a closed tube. The bottom is the closed end, so the water surface marks a node of air displacement.
As the water level drops, the change from one resonance event to the next one corresponds to the distance between adjacent nodes. This is one-half wavelength.
$$\Delta \mathcal{l}=\frac{1}{2}\lambda$$
$$\lambda=2\Delta \mathcal{l}=2(0.395m-0.125m)=0.540m$$
Now find the frequency.
$$f=\frac{v}{\lambda}=\frac{343m/s}{0.540m}=635\;Hz$$
