2 or 3

$\begin{aligned} A & = \frac {2+2^2+2^3+\cdots + 2^{3n-1}}{\frac 12 + \frac 1{2^2} + \frac 1{2^3}+ \cdots + \frac 1{2^{3n-1}}} = \frac {\frac {2\left(2^{3n-1}-1\right)}{2-1}}{\frac {\frac 12\left(1- \frac 1{2^{3n-1}}\right)}{1-\frac 12}} = 2^{3n} \\ B & = \frac {3+3^2+3^3+\cdots + 3^{2n-1}}{\frac 13 + \frac 1{3^2} + \frac 1{3^3}+ \cdots + \frac 1{3^{2n-1}}} = \frac {\frac {3\left(3^{2n-1}-1\right)}{3-1}}{\frac {\frac 13\left(1- \frac 1{3^{2n-1}}\right)}{1-\frac 13}} = 3^{2n} \end{aligned}$

Consider $\dfrac {\log A}{\log B} = \dfrac {\log 2^{3n}}{\log 3^{2n}} = \dfrac {n \log 2^3}{n \log 3^2} = \dfrac {\log 8}{\log 9} < 1$ . $\implies \log B > \log A$ , $\implies \boxed{B > A}$ .

Expand $A$ with $2^{3n-1}$ . Then we get: $\begin{aligned} A & =\dfrac{2^{3n-1}(2+2^2+2^3+\dots+2^{3n-1})}{2^{3n-2}+2^{3n-3}+\dots+2+1} \\ & =\dfrac{2^{3n}(1+2+4+\dots+2^{3n-2})}{2^{3n-2}+2^{3n-3}+\dots+2+1} \\ & = 2^{3n} \end{aligned}$

Similarly wif we expand $B$ with $3^{2n-1}$ and pull out $3^{2n}$ , we get $B=3^{2n}$ Now it is clear that $A=2^{3n}=8^n<9^n=3^{2n}=B$ Therefore $\boxed{B>A}$ .