Let the arithmetic progression have first term $a$ and common difference $d$.

The sum of the first $m$ terms of an AP is $S_m = \frac{m}{2}\left[2a + (m-1)d\right].$

Write $S_{3n}$ and $S_{2n}$: $S_{3n} = \frac{3n}{2}\left[2a + (3n-1)d\right],$ $S_{2n} = \frac{2n}{2}\left[2a + (2n-1)d\right] = n\left[2a + (2n-1)d\right].$

Given $S_{3n} = 3S_{2n}.$ Substitute the expressions: $\frac{3n}{2}\left[2a + (3n-1)d\right] = 3n\left[2a + (2n-1)d\right].$

Divide both sides by $3n$: $\frac{1}{2}\left[2a + (3n-1)d\right] = 2a + (2n-1)d.$

Multiply by $2$: $2a + (3n-1)d = 4a + 2(2n-1)d.$ $2a + (3n-1)d = 4a + (4n-2)d.$

Rearranging: $0 = 2a + (n-1)d,$ so $2a = -(n-1)d.$

Now compute $S_{4n}$: $S_{4n} = \frac{4n}{2}\left[2a + (4n-1)d\right] = 2n\left[2a + (4n-1)d\right].$

Using $2a = -(n-1)d$, $2a + (4n-1)d = -(n-1)d + (4n-1)d = 3nd.$ Hence, $S_{4n} = 2n(3nd) = 6n^2 d.$

Similarly, $S_{2n} = n\left[2a + (2n-1)d\right].$ Using $2a = -(n-1)d$, $2a + (2n-1)d = -(n-1)d + (2n-1)d = nd.$ Hence, $S_{2n} = n(nd) = n^2 d.$

Therefore, $\frac{S_{4n}}{S_{2n}} = \frac{6n^2 d}{n^2 d} = 6.$

So the correct option is: $\boxed{A}$