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

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

$E = \sigma A T^4$

For the spherical ball $A$, the given values are:
$A = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$
$T = 300 \text{ K}$
$\sigma \approx 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$

Substituting these values into the formula:

$E = (5.67 \times 10^{-8}) \times (20 \times 10^{-4}) \times (300)^4$

$E = 5.67 \times 10^{-8} \times 20 \times 10^{-4} \times 81 \times 10^8$

$E = 5.67 \times 1620 \times 10^{-4}$

$E \approx 0.918 \text{ J/s}$

The closest option provided is $0.94 \text{ J}$. (Note: Taking the Stefan-Boltzmann constant as $\sigma = 5.8 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$ yields exactly $0.9396 \approx 0.94 \text{ J/s}$).

Answer: $0.94 \text{ J}$