Here it is the simple calculation.

The problem is that of the propagation into a waveguide, so wavelength is not the same as for a wave traveling in the open-space, and depends on the mode involved. Each mode (field conformation) has its wavelength (and propagation speed, of course). Then the ideal situation is when there is only one mode propagating, so above a certain minimum frequency and below a certain other, which are a function of the diameter of the cavity.

Since the back of the can (closed side) imposes a zero to parallel electric field, positioning the pin at a distance equal to one quarter of the length lambda_g gives the best result (emitter antenna is into a point of maximum field strength).

Imposing the minimum and maximum frequencies to stay in between the mode and second mode cut-off:

a = d/2 > 1.83 / (2*pi*f_min*sqrt(mu0*epsilon0)) above the first mode cutoff

a = d/2 < 2.405 / (2*pi*f_max*sqrt(mu0*epsilon0)) below the second mode cutoff

Once the diameter is chosen (for wi-fi, 2.4 to 2.5 GHz, it is easy to find one satisfying the above conditions)

fTE_11 = 1.83 / (2*pi*a*sqrt(mu0*epsilon0)) first mode cutoff frequency

lambda_g = 1 / ( f*sqrt(mu0*epsilon0)*sqrt(1-(fTE_11/f)^2) )

mu0 and epsilon0 are the magnetic and electrical permeability constants of the air.

For a narrow bandwidth signal, f can be the central value, for wi-fi I'd choose the geometric average f = f_avg = sqrt(f_min*f_max).

In the example on my textbook it isn't mentioned, but IMO there is also an optimal can length, which should be

lambda_g/4 + N*lambda_g/2, with N at least 2 or 3. In that case, one could try to estimate the antenna gain considering it as a circular aperture antenna. I can't give the right formula at the moment, just can say that radiation goes as a Bessel function of the angle.

The maximum in the radiation diagram is in front of the opened side of the can, of course.

Since the reciprocity theorem (I think it is called so), the considerations made for the emission of radiation are true for receiving, too.

Hope it's worth. If the calculation seems complex, it's just me unable to explain, so just ask.