imc

Note that for all real numbers $k,$ we have $a^{2k}+a^{-2k}+2=(a^k+a^{-k}{)}^2,$ from which $$a^{2k}+a^{-2k}=(a^k+a^{-k}{)}^2-2.$$ We apply this result twice to get the answer: $$\begin{array}{rl}{a}^4+a^{-4}& =(a^2+a^{-2}{)}^2-2\\ & =[(a+a^{-1}{)}^2-2{]}^2-2\\ & =\boxed{194}.\end{array}$$