$ \underbrace{a^{a^{a^{\ldots^{a}}}}}_{1992}>1992=(\sqrt[1992]{1992})^{1992}=a^{1992} \Leftrightarrow \underbrace{a^{a^{a^{\ldots^{a}}}}}_{1991}>1992 \Leftrightarrow \dots \dots \Leftrightarrow a>1992$, which is impossible, so $ 1992$ is larger.

Fix integer $m>1$, let $A_0=1<m,\; a=m^{1/m},\; A_n = a^{A_{n-1}}$. Then $1\le A_0 < A_1 < m,\; A_n = \underbrace{a^{a^{a^{\ldots^a}}}}_{n}$ and $(1\le A_{n-1}<A_n<m)\implies (1\le A_n = a^{A_{n-1}}<a^{A_n}=A_{n+1}<a^m=m)$. Thus $1\le A_{n-1} < A_n < m$ for each positive integer $n$. In particular, $A_m < m$ with $m=1992$.