Let $a$ be first term and $d$ be common difference of this A.P.

Given $S_{3n}=3S_{2n}$

$\Rightarrow \frac{3n}{2}\left[2a+\left(3n-1\right)d\right]=3\times \frac{2n}{2}\left[2a+\left(2n-1\right)d\right]$

$\Rightarrow 2a+(3n-1)d=4a+(4n-2)d$

$\Rightarrow 2a+(n-1)d=0$

Now $\frac{S_{4n}}{S_{2n}}=\frac{\frac{4n}{2}\left[2a+\left(4n-1\right)d\right]}{\frac{2n}{2}\left[2a+\left(2n-1\right)d\right]}=\frac{2\left[2a+\left(n-1\right)d+3nd\right]}{\left[2a+\left(n-1\right)d+nd\right]}$

$=\frac{6nd}{nd}=6$