Current CMB photon number density?

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The current number of cosmic microwave background (CMB) photons per cubic meter can be estimated using the Stefan-Boltzmann law and Planck's law. The total energy density from the Stefan-Boltzmann law requires a temperature, while the mean photon energy is given as 2.70kT. The average photon energy of 2.7kT is confirmed to be applicable for calculating photon number density. The energy density in SI units is expressed as 7.56x10^-16T^4 in J/m^3. Understanding the relationship between energy density and photon energy is crucial for accurate calculations.
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Estimate the current number of cosmic microwave background (CMB) photons there are now, per cubic meter, given that the mean photon energy of a blackbody distribution is about 2.70kT.

Method

http://en.wikipedia.org/wiki/Planck_function" gives the energy density (i.e. energy per unit volume) per unit wavelength interval.

The http://en.wikipedia.org/wiki/Stefan-Boltzmann_law" is derived by integrating Planck's law, to give the total energy density (over all wavelengths).

The photon number density (photons per cubic metre) can be found by (a) taking the total energy density (from the Stefan-Boltzmann Law) and dividing it by (b) the typical photon energy.

Problem

It looks like the Stefan–Boltzmann law should help, but it requires a temperature, T, whereas I'm given the mean photon energy in kT.

Planck's law includes a term for kT, but results in the energy density per unit wavelength.

I'm given the mean photon energy, 2.70kT. Is this necessarily the same as the typical photon energy?

I'm obviously missing a crucial next step, can anyone help?
 
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The average photon energy is in fact 2.7kT (I know it sounds a bit high). The number of BB photons in a thermalized sample space (BB cavity) multiplied by 2.7 kT is the total photon energy of that sample space. This relationship can be inverted in the expected way.

In SI units, the energy density is 7.56x10-16T4 in J/m3.
 

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