How is temperature of space calculated? wikipedia says:

The CMB has a thermal black body spectrum at a temperature of
  (T=) 2.72548±0.00057 °K. The spectral radiance dEν/dν peaks at 160.23 GHz

Could you explain how is T= 2.72 K obtained? is it an average, is it an integral or what? Is the thermal spectrum fixed at each temp?
If we multiply 2.72 by $k_{B}$ (= 2.83 x 10^10) we get a frequency of 50.45 GHz, which is less than 1/3 of the given frequency, can you explain how is that value calculated?
 A: The CMB, as wikipedia says, has a black body spectrum. As you may or may not know, from the black body spectrum, one can obtain the temperature of the black body through Wien's displacement law, which states that:
$$
\lambda_{max} = \frac{b}{T}
$$
Where $b$ is a constant. This means that the wavelength ($\lambda_{max}$) for which the spectrum reaches its maximum, is inversely proportional to the temperature. Plugging in the $\lambda_{max}$ obtained from the CMB spectrum, you should get around $2.73K$. The peak frequency is related to the peak wavelength by: $\nu_{max}=c/\lambda_{max}$.

This is what black body radiation spectra look like for different temperatures.
A: The CMB has a thermal black body spectrum at a temperature of 2.72548±0.00057 K. The spectral radiance $dE_\nu/d\nu$ peaks at 160.23 GHz. Alternatively, if spectral radiance is defined as $dE_\lambda/d\lambda$, then the peak wavelength is 1.063 mm (282 GHz).
The difference is in the definition of spectral radiance. Radiance per unit of frequency and radiance per unit of wavelength do not match, but happen in different parts of the spectrum.
See: Cosmic Microwave Background
According to Plank's law
$$E_\nu(T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{kT}} - 1}$$
Differentiating by $\nu$ gives the peak of $dE_\nu/d\nu=0$ at
$$(3-x)e^x-3=0$$
that easily solves numerically for $x\equiv\dfrac{h\nu}{kT}=2.82143937$...
Using the constants
$$h=6.62607004\cdot 10^{-34}$$
$$k=1.3806485\cdot 10^{-23}$$
We get
$$\nu_{max}=T\cdot 58.789\,GHz$$
Which for $T=2.725\,K^o$ yields $\nu_{max}=160.23\,GHz$ at $\lambda=1.871\,mm$
