Discrepancy of $4π/c$ for radiated intensity in a blackbody radiation Plancks radiation law is sometimes given as
$B(\nu, T)=\frac{8π h \nu^{3}}{c^{3}} \frac{1}{e^{\frac{h \nu}{k_{\mathrm{B}} T}}-1}$ as is shown in the plot

But Wikipedia says:

...the spectral radiance of a body for frequency $ν$ at absolute temperature $T$ is given by
$B(\nu, T)=\frac{2 h \nu^{3}}{c^{2}} \frac{1}{e^{\frac{h \nu}{k_{\mathrm{B}} T}}-1}$

So there appears to be a discrepancy of $4π/c$ in the Wikipedia and the formula given on the plot.
Why is this so?
(The plot maybe takes into account all of the radiation while as the Wikipedia one is talking about radiation per unit solid angle,but that doesn't explain the extra factor $c$)
The duplicate linked question which is assumed to be be the same as this one are different. That question is about the wavelength and frequency form of Plancks law but this one isn't.
 A: The extra factor of $4\pi/c$ turns the spectral radiance (or specific intensity) into an energy density with units of energy per unit volume per unit frequency interval. There is an implicit integration over $4\pi$ steradians because blackbody radiation is isotropic. Dividing by $c$ is what turns a radiative flux (power per unit area) into an energy density (energy per unit volume).
Consider a bundle of light rays of specific intensity $I(\nu)$ (i.e. power per unit area, per unit frequency per unit solid angle) heading into a circular aperture. The amount of energy that passes through that aperture is
$$dE(\nu) = I A\ d\Omega\ dt\  ,$$
where $A$ is the cross-sectional area and $d\Omega$ is the solid angle subtended by the light ray bundle.
In time $dt$, that energy fills a space of volume $Ac\ dt$. Thus the energy density is
$$ du(\nu) = \frac{I}{c}\ d\Omega\ .$$
But for blackbody radiation, $I(\nu)=B_{\nu}$ and is isotropic and so it is straightforward to integrate over all solid angle to get
$$ u(\nu) = \frac{4\pi}{c} B_{\nu}\ .$$
I have no idea why $u(\nu)$ should be labelled "radiated intensity", since it does not have units of intensity.
A: If you are familiar with statistical mechanics, write down the density of states between E and E+dE in a box with $$n=\sqrt{n_{1}^2+n_{2}^2+n_{3}^2}$$
$$g(E)dE = 2\frac{4\pi n^2}{8}dn$$
Facfor of two comes from the fact that for any n component there will be two photon states. Wavelengths are related with
$$\lambda = \frac{2L}{n}$$
To get dE, one can write above equation as $$E = \frac{hc} {\lambda} = \frac{hc}{2L}n$$
From the density of states
$$g(E)dE = 2\frac{4\pi n^2}{8}dn =\frac{8\pi L^3 E^2}{(hc)^3}dE$$
Finally integrate this over all possible energies, it will give you
$$\frac{8\pi h \nu^3}{c^3} \frac{1}{e^{h\nu /kT} - 1}$$
Edit: make sure that you divide it by the volume of the box since it refers to spectral energy density
