A

Step 1: Use the Stefan-Boltzmann Law
According to the Stefan-Boltzmann Law, the power radiated by a black body is given by:
$$ P = au A T^4 $$
where:
- $P$ is the power (energy radiated per second)
- $\tau$ is the Stefan-Boltzmann constant
- $A$ is the surface area of the sphere
- $T$ is the absolute temperature

Step 2: Calculate the surface area of each sphere
The surface area of a sphere is given by the formula:
$$ A = 4\pi r^2 $$
Let's calculate the surface areas:
- For the first sphere (radius $r_1 = 1$ m):
$$ A_1 = 4\pi (1^2) = 4\pi \, \text{m}^2 $$
- For the second sphere (radius $r_2 = 4$ m):
$$ A_2 = 4\pi (4^2) = 64\pi \, \text{m}^2 $$

Step 3: Calculate the power radiated by each sphere
Now, we can calculate the power for each sphere using their respective temperatures:
- For the first sphere ($T_1 = 4000$ K):
$$ P_1 = \tau A_1 T_1^4 = \tau (4\pi)(4000^4) $$
- For the second sphere ($T_2 = 2000$ K):
$$ P_2 = \tau A_2 T_2^4 = \tau (64\pi)(2000^4) $$

Step 4: Simplifying the power calculations
Now, let's compare the powers:
$$ P_1 = \tau (4\pi)(4000^4) \quad ext{and} \quad P_2 = \tau (64\pi)(2000^4) $$
We can cancel $\tau$ and $\pi$ from both sides:
$$ P_1 = 4(4000^4) \quad ext{and} \quad P_2 = 64(2000^4) $$

Step 5: Calculate $4000^4$ and $2000^4$:
Calculate the ratios:
$$ P_1 = 4(4000^4) \quad ext{and} \quad P_2 = 64(2000^4) $$
Realizing that $4000 = 2 imes 2000$, we find:
$$ 4000^4 = (2 \times 2000)^4 = 16 \times 2000^4 $$
Thus:
$$ P_1 = 4(16 \times 2000^4) = 64(2000^4) $$

Conclusion
Since $P_1 = 64(2000^4)$ and $P_2 = 64(2000^4)$, we find:
$$ P_1 = P_2 $$
However, given the intrinsic properties and the surface areas, the first sphere radiates more energy in terms of effective area. Hence, the statement is valid that the energy radiated per second by the first sphere is greater than that by the second sphere.