Solution:

To solve this, we use Stefan-Boltzmann Law for radiative heat transfer:

\[
P = \sigma A (T^4 - T_s^4)
\]

Where:
- \(P\) is the power (rate of energy loss),
- \(\sigma\) is the Stefan-Boltzmann constant,
- \(A\) is the area of the body,
- \(T\) is the temperature of the body in kelvins,
- \(T_s\) is the temperature of the surroundings in kelvins.

 Step 1: Convert temperatures to kelvins
- Initial temperature of the black body: \(727^\circ C = 727 + 273 = 1000\,K\)
- Surrounding temperature: \(227^\circ C = 227 + 273 = 500\,K\)
- New temperature of the black body: \(1227^\circ C = 1227 + 273 = 1500\,K\)

 Step 2: Ratio of power losses
Let the power at \(T_1 = 1000\,K\) be \(P_1 = 20\,W\). The new power at \(T_2 = 1500\,K\) is \(P_2\). Using the Stefan-Boltzmann law:

\[
\frac{P_2}{P_1} = \frac{T_2^4 - T_s^4}{T_1^4 - T_s^4}
\]

Substitute the values:

\[
\frac{P_2}{20} = \frac{1500^4 - 500^4}{1000^4 - 500^4}
\]

Step 3: Simplify the powers of temperatures
\[
\frac{P_2}{20} = \frac{1500^4 - 500^4}{1000^4 - 500^4}
\]

You can approximate the values:

\[
P_2 = 20 \times \frac{(1500^4 - 500^4)}{(1000^4 - 500^4)}
\]

After simplifying, you'll find:

\[
P_2 = \frac{320}{3} \text{ watts}
\]

Thus, the rate of energy loss at 1227°C is \(\frac{320}{3}\,W\).