Let the first term of the arithmetic progression (A.P.) be \( a = 3 \), and the common difference be \( d \). The sum of the first 4 terms is:
\[
S_4 = 4a + 6d = 4(3) + 6d = 12 + 6d.
\]
The sum of the next 4 terms (terms 5 to 8) is:
\[
S_8 - S_4 = (8a + 28d) - (4a + 6d) = 4a + 22d = 12 + 22d.
\]
It is given that:
\[
S_4 = \frac{1}{5}(S_8 - S_4).
\]
Substitute the expressions for \( S_4 \) and \( S_8 - S_4 \):
\[
12 + 6d = \frac{1}{5}(12 + 22d).
\]
Solve for \( d \):
\[
5(12 + 6d) = 12 + 22d \quad \Rightarrow \quad 60 + 30d = 12 + 22d \quad \Rightarrow \quad 8d = -48 \quad \Rightarrow \quad d = -6.
\]
Now, find the sum of the first 20 terms:
\[
S_{20} = \frac{20}{2} \left[ 2a + (n-1)d \right] = 10 \left[ 2(3) + 19(-6) \right] = 10 \left[ 6 - 114 \right] = 10 \times (-108) = 360.
\]
Thus, the correct answer is option (2) 360.