$\displaystyle 2^{3^{2^{3^{2^3}}}}$ or$\displaystyle 3^{2^{3^{2^{3^2}}}}$
Recall that:
$\displaystyle (x^a)^b = x^{ab}$
So simply multiply all of the exponents out to simplify your expression:
$\displaystyle
2^{3^{2^{3^{2^3}}}} = 2^{108}
$
and:
$\displaystyle
3^{2^{3^{2^{3^2}}}} = 3^{72}
$
Now simply punch the two into your calculator.
$\displaystyle 2^{3^{2^3}}=2^{3^8}=2^{6561}$
$\displaystyle 3^{2^{3^2}}=3^{2^9}=3^{512}$
$\displaystyle 2^{6561}$ is a very much bigger number than $\displaystyle 3^{512}$, and since $\displaystyle \ln{3}>\ln{2}$, therfore $\displaystyle 2^{6561}\ln{3}$ is a very much bigger number than $\displaystyle 3^{512}\ln{2}$. On the other hand, $\displaystyle \ln{(\ln{2})}<\ln{(\ln{3})}$ but the difference is minuscule in comparison. Therefore, I think we can conclude that
$\displaystyle 2^{6561}\ln{3}+\ln{(\ln{2})}\ >\ 3^{512}\ln{2}+\ln{(\ln{3})}$
that is to say,
$\displaystyle 2^{3^{2^{3^{2^3}}}}\ >\ 3^{2^{3^{2^{3^2}}}}$