B

Step 1: Use Stefan-Boltzmann Law, which states that the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature:
$$E = \sigma T^4$$
where E is the emissive power, T is the absolute temperature in Kelvin, and \sigma is the Stefan-Boltzmann constant.
Step 2: Convert temperatures from Celsius to Kelvin:
- For 127ºC: \( T_1 = 127 + 273.15 = 400.15 \, K \)
- For 527ºC: \( T_2 = 527 + 273.15 = 800.15 \, K \)
Step 3: Calculate the rate of radiation at both temperatures:
\( E_1 = \sigma (400.15)^4 \) and \( E_2 = \sigma (800.15)^4 \).
Step 4: Find the ratio of the rates of radiation:
\( \frac{E_2}{E_1} = \frac{(800.15)^4}{(400.15)^4} = 16 \)
Therefore, the rate of radiation at 527ºC is 16 times greater than that at 127ºC.
Therefore, the correct answer is the option that indicates this factor.