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

According to the Stefan-Boltzmann law, the total radiation energy emitted per second by a blackbody surface is given by:

$E = \sigma A T^4$

For the inner surface of the hollow spherical shell $B$:
Area, $A_B = 80 \text{ cm}^2 = 80 \times 10^{-4} \text{ m}^2$
Temperature, $T = 300 \text{ K}$
Stefan-Boltzmann constant, $\sigma \approx 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$

Substituting the given values into the formula:

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

$E_B = 5.67 \times 10^{-8} \times 80 \times 10^{-4} \times 81 \times 10^8$

$E_B = 5.67 \times 80 \times 81 \times 10^{-4}$

$E_B = 3.674 \text{ J/s}$

Rounding to the nearest given option, we get approximately $3.8 \text{ J}$.
(Note: If $\sigma$ is approximated as $5.8 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$ as is sometimes done to match specific options, $E_B = 3.76 \text{ J/s} \approx 3.8 \text{ J/s}$).

Answer: $3.8 \text{ J}$