A spherical ball A having a surface area of 20 cm^2 is positioned...

Let $F_{AB}$ be the fraction of energy emitted by sphere $A$ that falls on sphere $B$. Since $A$ is completely enclosed by $B$, all radiation emitted by $A$ falls on $B$, giving $F_{AB} = 1$.

Using the reciprocity theorem for thermal radiation:

$A_A F_{AB} = A_B F_{BA}$

$F_{BA} = \dfrac{A_A}{A_B} F_{AB} = \dfrac{20}{80} \times 1 = 0.25$

The fraction of energy emitted by $B$ that falls back onto its own inner surface is:

$F_{BB} = 1 - F_{BA} = 1 - 0.25 = 0.75$

The total energy emitted per second by the inner surface of $B$ (acting as a blackbody) is given by Stefan-Boltzmann's law:

$E_B = \sigma A_B T^4$

The energy falling back onto $B$ per second is:

$E_{BB} = F_{BB} E_B = 0.75 \sigma A_B T^4$

Substituting the given values ($\sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4$, $A_B = 80 \times 10^{-4} \text{ m}^2$, $T = 300 \text{ K}$):

$E_{BB} = 0.75 \times 5.67 \times 10^{-8} \times 80 \times 10^{-4} \times (300)^4$

$E_{BB} = 5.67 \times 10^{-8} \times 60 \times 10^{-4} \times 81 \times 10^8$

$E_{BB} = 5.67 \times 60 \times 81 \times 10^{-4}$

$E_{BB} = 5.67 \times 0.486 \approx 2.756 \text{ J/s}$

Rounding to one decimal place, the energy falling back onto the surface is $2.8 \text{ J}$.