What is Size of Photon?

A photon is a unit ("quantum") of excitation of the quantum electromagnetic field. Thinking roughly of the quantum field as a vast collection of quantum harmonic oscillators, each oscillator corresponding to a mode of vibration of the field, we specify the quantum field's state by stating how many quantums above the QHO ground state each mode oscillator is in (recall that a quantum harmonic oscillator has equispaced energy levels of even energy spacing $h\,\nu$ with ground state energy $\frac{1}{2}\,h\,\nu$). The one and only physical entity in this picture is the quantum field, the "photons" are just units used to name the state of mode oscillators, just as Euros or Dollars or Rupees or Yens might be used to name the state of your bank account. The field doesn't even have to have a certain number of photons in each oscillator: being a quantum object, it can be in a linear quantum superposition of states with definite photon numbers (superposition of Fock states).

So one can no more ask what the physical extent of a photon is any more than one can ask what the physical extent of the integer $1$ is. I would commend the Physics SE question "Which is more fundamental, Fields or Particles?" and user DanielSank's answer in particular to find out more about these ideas.

However, one can meaningfully ask for characteristic sizes of regions significantly influenced by the electromagnetic field in a pure one-photon state. As with the electron field, we can delocalize the disturbance arbitrarily: a one photon state that is a momentum eigenstate is theoretically delocalized over all space. In general, one photon states are extremely hard to confine to regions smaller than about a wavelength. The electromagnetic field can in special circumstances be confined to smaller regions, but it then becomes evanescent and in any case this doesn't happen in freespace: interaction with matter is needed so that we aren't really talking about pure photons anymore, but rather superpositions of EM and matter excitations.

Continuing the line of thought that defines size in term of crossection with respect to a relevant process requires one to clarify the interaction with respect to which you want to know photon's size.

To my taste, a natural crossection to look for would be scattering of photons on themselves in the Standard model vacuum, as described here: https://en.m.wikipedia.org/wiki/Two-photon_physics

A quick search of recent literature turns this paper: https://arxiv.org/abs/1106.0592 where the (differential) crossection is given in terms of the fine-structure constant (QED coupling constant) $\alpha$ and photon frequency $\omega$:

$$\frac{d\sigma}{d\Omega}=\frac{1}{(6\pi)^2}\frac{\alpha^4} {(2\omega)^2}(3+2\cos^2\theta +\cos^4\theta)$$

One just needs to integrate it over angles and put in proper units (restore the factor $c^2$ lost to $c=1$ convention and substitute $\omega = 2 \pi c /\lambda$) to get the total scattering crosssection in terms of photon-wave length squared and the fourth power $\alpha$: $$\Sigma= 2 \pi \int_0^{\pi} \frac{d\sigma}{d\Omega}\sin \theta \, d \theta=\frac{29 \, \alpha ^4 \, c^2}{270 \, \pi \, \omega ^2} = \frac{29 \,\alpha ^4 \lambda ^2}{1080 \, \pi ^3} \, .$$

Hence a photon in the eyes of other photons with the same wave-lengths is equivalent to an opaque disk of diameter $$d \equiv \sqrt{\frac{4 \Sigma}{\pi}} =\sqrt{\frac{29}{30}} \frac{\alpha^2}{3 \pi^2} \lambda= 1.768 \cdot 10^{-6} \, \lambda \, .$$.

The answer is obviously proportional to the wavelength (there is no other lengths in the problem as long as $\lambda$ is much smaller than the Compton wavelength of the electron, which is the limit considered in the paper). However, the proportionality constant is very small owing to the small value of $\alpha$ and the fact that light-light scattering is a fourth order QED process.