Deciding the set of all Turing machines that halt in at most $k|x|$ steps $\forall x \in \Sigma^*$
The language is actually decidable, but the proof is quite involved, and uses crossing-sequence arguments. See this paper for details.

If you change $200|x|$ to a faster growing function, such as $|x|\log |x|$ (plus some constant), then the language becomes undecidable by standard reduction arguments.