Solution

The energy density of thermal radiation (equilibrium black-body radiation) in a cavity is related to the radiant emittance by the formula:
$$u = \frac{4\sigma}{c} T^4$$
where:

* $\sigma = 5.67 \cdot 10^{-8} \text{ W/(m}^2 \cdot \text{K}^4\text{)}$ is the Stefan-Boltzmann constant,
* $c = 3 \cdot 10^8 \text{ m/s}$ is the speed of light,
* $T$ is the absolute temperature.

Let us substitute the numerical values of the constants to obtain the final calculation formula:
$$u = \frac{4 \cdot 5.67 \cdot 10^{-8}}{3 \cdot 10^8} T^4 = 7.56 \cdot 10^{-16} T^4 \text{ J/m}^3$$
(Note: the numerical coefficient $7.56 \cdot 10^{-16}$ itself has units of $\text{J}/(\text{m}^3 \cdot \text{K}^4)$).