Ratio of Radiated Power with Parallel Black Body Radiation Shields

In the initial configuration (Fig. 1), the net radiated power per unit area transferred between the two black body plates $\mathrm{P}$ and $\mathrm{Q}$ is given by Stefan-Boltzmann law: $W_0 = \sigma \left(T_\mathrm{P}^4 - T_\mathrm{Q}^4\right)$

In the second configuration (Fig. 2), two identical plates are inserted between $\mathrm{P}$ and $\mathrm{Q}$. Let the steady-state temperatures of the two intermediate plates be $T_1$ and $T_2$ such that $T_\mathrm{P} > T_1 > T_2 > T_\mathrm{Q}$.

In the steady state, the net heat flux $W_S$ between each pair of adjacent plates must be equal: $W_S = \sigma \left(T_\mathrm{P}^4 - T_1^4\right) = \sigma \left(T_1^4 - T_2^4\right) = \sigma \left(T_2^4 - T_\mathrm{Q}^4\right)$

Summing the heat transfer equations across the three regions: $3W_S = \sigma \left(T_\mathrm{P}^4 - T_1^4\right) + \sigma \left(T_1^4 - T_2^4\right) + \sigma \left(T_2^4 - T_\mathrm{Q}^4\right) = \sigma \left(T_\mathrm{P}^4 - T_\mathrm{Q}^4\right)$

Substituting $W_0 = \sigma \left(T_\mathrm{P}^4 - T_\mathrm{Q}^4\right)$: $3W_S = W_0 \implies \frac{W_0}{W_S} = 3$