Irradiation, Radiosity, and Emissive Power:
Irradiation is the incident radiant energy per unit area on a surface. It is the amount of energy that is received by the surface from all sources. Radiosity, on the other hand, is the total radiant energy leaving the surface per unit area and is equal to the sum of the reflected and emitted energy. Emissive power is the radiant energy emitted per unit area by the surface.

Given:
Irradiation (I) = 20
Radiosity (J) = 12
Emissive Power (E) = ?

Relationship between Irradiation, Radiosity, and Emissive Power:
The relationship between these three parameters can be defined by the following equation:

J = (1 - ρ) * I + E

Where:
J is the radiosity,
I is the irradiation,
E is the emissive power, and
ρ is the reflectivity of the surface.

Finding the Emissive Power:
We need to find the emissive power (E) of the surface. Since the surface is opaque, it does not reflect any energy, so the reflectivity (ρ) is zero. Therefore, the equation becomes:

J = (1 - 0) * I + E
J = I + E

Substituting the given values:
12 = 20 + E

Simplifying the equation:
E = 12 - 20
E = -8

Understanding Negative Emissive Power:
A negative emissive power does not make physical sense because emissive power represents the radiant energy emitted by the surface, which should always be positive. Therefore, there must be a mistake in the given values.

Correcting the Mistake:
Since the emissive power cannot be negative, the given values must be incorrect. Let's assume the correct value for the emissive power is 8.

J = I + E
12 = 20 + 8

Simplifying the equation:
8 = 28

This equation does not hold true, which means our assumption for the emissive power was incorrect.

Correct Answer:
Since the given values do not lead to a valid solution, it is not possible to determine the emissivity of the surface. Therefore, the correct answer is that the emissivity cannot be determined with the given information.