Some simple logic can help us solve this question without too much work. Let us denote $f_1(x) = x$, $f_2(x) = x^x$, $f_3(x) = x^{x^x}$ and so on. We want to solve $$ \lim_{n \rightarrow \infty} f_n(x) = 2. $$ In the limit of $n \rightarrow \infty$, then it must be the case that $$ \lim_{n \rightarrow \infty} f_n(x) = \lim_{n \rightarrow \infty} f_{n-1}(x) $$ that is, adding or removing one $x$ from the chain of the exponents should give the same result. It is clear that $$ f_n(x) = x^{f_{n-1}(x)} = 2 $$ and since we also see that $f_{n-1}(x) = 2$, then we have $$ x^2 = 2 $$ or $$ x = \sqrt{2}. $$ To three decimal places, the answer is $1.414$.