What is Size of Photon? Is there any size of photon if so what is it?
And also which particle had smallest size / radius / volume considering all of the matter.
 A: 
Is there any size of photon if so what is it?

The photon is an elementary particle among the others which form a basis for the standard model of particle physics.


The Standard Model of elementary particles (more schematic depiction), with the three generations of matter, gauge bosons in the fourth column, and the Higgs boson in the fifth.

The model encapsulates all the experimental data very successfully, and the size of these particles is considered zero, they are point particles. 

And also which particle had smallest size / radius / volume considering all of the matter.

A composite particle, like a proton which is made up from three quarks and their dynamic exchanges, has a definite size, given the energy of the probing interaction.
Elementary particles are point particles.
String theory which is trying to extend the standard model and unify it with gravitation  hypothesizes that elementary particles are vibrations on a one dimensional string, whose dimension is of order of the planck length, 16x10^-36meters, a very small length not measurable experimentally.
A: 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.
A: 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.
A: For particles, "size" is often defined by the statistics of scattering experiments.  A reasonable way to measure the length of a photon is to first provide a source of a lot of identical photons, then set up a Michaelson interferometer with a variable-length arm.  The interferometer can then be used to measure the maximum path length difference that allows formation of an interference pattern.  That is equivalent to half of the photon's length.  The width of a photon can be measured similarly: in a Michaelson interferometer, include mirrors or prisms to flip the beam in one arm so that one beam is superimposed on a mirror image of itself.  The width of the region over which interference fringes is observable is equivalent to half of the photon's width.  Note that the size of a photon per this definition is not a constant; it depends on the type of light source, and on the geometry of the interferometer optics.  
If an analogous method is used to measure the size of an electron, a similar result is obtained - because this measures the size of the wave function.  On the other hand, a proton has a finite radius that can be measured by scattering experiments (e.g., proton-proton scattering).  Photon-photon scattering does not occur in any currently obtainable conditions, and electron-electron scattering, so far, has not revealed a finite radius.
Bottom line: particle size depends on the definition of size, which boils down to measurement method.  I personally am comfortable with saying that a photon's size is the size of its wavefunction.
A: To add to the previous answers and clarify, if you accept that when we say photon we are really referring to a disturbance in an EM field then the answer is the photon doesn't have volume because waves don't have volume.
I think this is a confusing idea for some, because classically when we talk about a wave we say there is an underlying particle that interacts through some force and the transfer in energy through this medium of particles by this force forms the wave.  But with photons, the claim is there is no medium.
This really brings up the question how does a photon propagate if there is no medium. One way of answering this question is to ask what is the mechanism by which photons interact with each other, discussed a bit here. Hopefully that provides some more context to the above answers. 
