Step-by-step Solution

1. Analyze the initial configuration (Fig. 1)

In the first case, we have two identical parallel plates, P and Q, which are perfect black bodies. They are at constant absolute temperatures $T_P$ and $T_Q$ respectively, with $T_P > T_Q$.

According to the Stefan-Boltzmann law, the power radiated per unit area by a black body at temperature $T$ is given by $E = \sigma T^4$, where $\sigma$ is the Stefan-Boltzmann constant.

The net power transferred per unit area from the hotter plate P to the colder plate Q, denoted by $W_0$, is the difference between the power radiated by P and absorbed by Q, and the power radiated by Q and absorbed by P. Since they are large parallel plates, we assume they exchange radiation only with each other.

$W_0 = (\text{Power radiated by P}) - (\text{Power radiated by Q})$ $W_0 = \sigma T_P^4 - \sigma T_Q^4 = \sigma (T_P^4 - T_Q^4) \quad \cdots(1)$

2. Analyze the final configuration (Fig. 2)

In the second case, two more identical plates are introduced between P and Q. Let's call them R and S. The arrangement is P, R, S, Q.

In the steady state, the intermediate plates R and S will attain constant temperatures, let's say $T_R$ and $T_S$. Since $T_P > T_Q$, the temperatures will be in the order $T_P > T_R > T_S > T_Q$.

In steady state, the net rate of heat flow per unit area, $W_S$, must be the same through each gap between adjacent plates.

3. Set up equations for the steady state

The net power transferred per unit area is constant throughout the system:

• Between plates P and R: $W_S = \sigma (T_P^4 - T_R^4) \quad \cdots(2)$

• Between plates R and S: $W_S = \sigma (T_R^4 - T_S^4) \quad \cdots(3)$

• Between plates S and Q: $W_S = \sigma (T_S^4 - T_Q^4) \quad \cdots(4)$

4. Solve for $W_S$

We can rearrange equations (2), (3), and (4) to express the temperature differences (in terms of $T^4$) in terms of $W_S$:

$T_P^4 - T_R^4 = \frac{W_S}{\sigma}$ $T_R^4 - T_S^4 = \frac{W_S}{\sigma}$ $T_S^4 - T_Q^4 = \frac{W_S}{\sigma}$

To find $W_S$ in terms of the known temperatures $T_P$ and $T_Q$, we can add these three equations. This will eliminate the unknown intermediate temperatures $T_R$ and $T_S$.

$(T_P^4 - T_R^4) + (T_R^4 - T_S^4) + (T_S^4 - T_Q^4) = \frac{W_S}{\sigma} + \frac{W_S}{\sigma} + \frac{W_S}{\sigma}$

Canceling the intermediate terms on the left side:

$T_P^4 - T_Q^4 = 3 \frac{W_S}{\sigma}$

Solving for $W_S$:

$W_S = \frac{\sigma (T_P^4 - T_Q^4)}{3} \quad \cdots(5)$

5. Calculate the required ratio $\frac{W_0}{W_S}$

Now we can find the ratio using the expressions for $W_0$ from equation (1) and $W_S$ from equation (5).

$\frac{W_0}{W_S} = \frac{\sigma (T_P^4 - T_Q^4)}{\frac{\sigma (T_P^4 - T_Q^4)}{3}}$

$\frac{W_0}{W_S} = 3$

Thus, the required ratio is 3.