First, let's understand what it means for a state to be degenerate. In quantum mechanics, a degenerate state refers to a state with the same energy as another state. This means that multiple states can have the same energy value.

To determine the degeneracy of the first excited state of the 3-dimensional second box, we need to consider the possible energy levels and their corresponding wave functions.

1. Energy levels in a 3-dimensional box:
In a 3-dimensional box, the energy levels are given by the equation:

E = (n₁² + n₂² + n₃²) * (π²h² / 2mL²)

Here, n₁, n₂, and n₃ are the quantum numbers along the x, y, and z directions, respectively. h is Planck's constant, m is the mass of the particle, and L is the length of each side of the box.

2. First excited state:
The first excited state corresponds to the energy level with the next highest energy value compared to the ground state. In this case, we can consider the first excited state to have n₁ = 1, n₂ = 1, and n₃ = 1.

Plugging these values into the energy equation, we find:

E₁ = (1² + 1² + 1²) * (π²h² / 2mL²)

3. Degeneracy of the first excited state:
To determine the degeneracy of the first excited state, we need to consider if there are any other possible combinations of quantum numbers that give the same energy value.

For the first excited state, we can have the following combinations of quantum numbers that also result in E₁:

- n₁ = 1, n₂ = 2, n₃ = 0
- n₁ = 2, n₂ = 1, n₃ = 0
- n₁ = 0, n₂ = 1, n₃ = 2
- n₁ = 1, n₂ = 0, n₃ = 2
- n₁ = 2, n₂ = 0, n₃ = 1
- n₁ = 0, n₂ = 2, n₃ = 1

These combinations give the same energy value E₁ as the original state. Therefore, the first excited state is 3-fold degenerate.

Therefore, the correct answer is option 'C' - the first excited state is 3-fold degenerate.