quantum mechanics - What's the relationship between the energy density of a black-body and its radiant exitance?

Through a bit calculation we can derive that in a cavity, the energy density $$u(f,T)=\overline{E(f)}\times G(f)=\frac{8\pi h}{c^3}\frac{f^3}{e^{h\nu /kT}-1}$$ If we take the integral over all frequency, we get $$U(T)=\frac{8πh}{c^3}\frac{(kT)^4}{h^3}{\frac{π^4}{15}}=C_{onst}T^4$$ And Stefan-Boltzmann Law claims that for a perfect black-body $$j^*=\sigma T^4$$ where $j^*$ is the radiant exitance, which is defined as the total energy radiated per unit surface area of a black body across all wavelengths per unit time.

And it just so happens that $\frac{\sigma}{C_{onst}}=\frac{c}{4}$, why is that?

P.S. The professor told me to refer to some thermodynamics book, where a more general case is discussed. But we don't have that book in our library and the professor's now out of town xD.