What is the speed of a photon in water? What is the speed of light in water? The speed of light in vacuum, divided by the index of refraction for water. And what is the speed of a photon in water?
 A: For a plane wave there are two velocities, the phase velocity and the group velocity. We define these in terms of the angular velocity:
$$ \omega = 2\pi f $$
and the wave vector:
$$ k = \frac{2\pi}{\lambda} $$
Then the phase velocity is given by:
$$ v_p = \frac{\omega}{k} $$
and the group velocity by:
$$ v_g = \frac{d\omega}{dk} $$
For light moving in a vacuum the speed of light is constant so $\omega/k = c$ and $d\omega/dk = c$. Hence both the group and phase velocities are equal to $c$.
However when the light is passing through a medium things are more complicated. We define the refractive index as the ratio of the phase velocities, so in a medium with refractive index $n$ the phase velocity is simply:
$$ v_p(k) = \frac{c}{n(k)} $$
Note that I've written the refractive index as a function of the wavevector, $n(k)$. That is because in general the refractive index is not a constant but depends on the wavelength of the light. This means that when we differentiate to get the group velocity the result will not be the same as the phase velocity. In practice the difference is usually small since the refractive index changes relatively slowly with wavelength, but it is large enough to be easily measurable. Indeed the variation of the refractive index with wavelength is what causes the dispersion seen when light is refracted through a prism.
The reason I mention this is because the velocity of the photons is the group velocity not the phase velocity, so it will be slightly different to $c/n$. To calculate it precisely you need to know how the refractive index varies with wavelength. This is different for different dielectrics so you need to determine it experimentally.
A: $\let\lam=\lambda \let\om=\omega \def\ns#1#2{#1_{\mathrm#2}} 
 \def\vp{\ns vp} \def\vg{\ns vg}$
The idea of giving a different speed to a photon when it propagates
through a medium encounters several difficulties. To tell it shortly, it
would be a peculiar kind of particle from the point of view of SR.
What is photon's energy? And its momentum? If you take $E=\hbar\om$ and
$p=h/\lam$ you get $E/p=\vp$. Try to compute a mass:
$$m^2 c^4 = E^2 - c^2 p^2 = (\vp^2 - c^2)\,p^2 < 0$$
if $\vp<c$.
It would be better not to change photon properties from those holding
in vacuum, and give an alternative explanation for reflection,
refraction, and also for its apparent propagation delay. This is not
to say that @Nobody recognizeable's proposal may work. There is no
explanation for a "stop" within a molecule, for precisely the time
required to give right (reduced) speed.
A much better explanation can be found in Feynman's QED (pages
101-109). He leaves photon propagate as in vacuum, but takes into
account its scattering from atoms. The scattered amplitude is 90° out
of phase wrt the incoming one, and superposition of both amplitudes
causes a phase delay which we macroscopically interpret as due to a
reduced speed.
Feynman also explains by the same mechanism the apparent reflection of
photons on the surface of a transparent medium. The reflected wave is
just the coherent superposition of photon's scattered amplitudes by
atoms in the medium's bulk.
A: Well when the photon moves the speed of photon even in water is same as speed of light as vacuum ie $3\times10^8\ \rm m/s$ . Actually in water or in other mediums photons are captured by water molecules for a short amount of time (ie $10^{-11}$ seconds) and then released back. So the light appears to move slower as a whole .  So to your question I'll say when the photon moves it moves with speed of light in vacuum.  But when it is captured it does not move so for that odd amount of time the photon's velocity is zero. So the higher the refractive index higher the capturing time . And the speed of light or more precisely the average speed of light in water is speed of light in vacuum }/{refractive index of water} =$\frac{3\times10^8}{1.33} = 2.255 \times 10^8 \ \rm m/s$
