You can reduce the halting problem to your problem. Given a Turing machine $M$ and input $z$, let us design a new machine $N$, which on input $y$, first runs $M$ on input $z$ for $|y|$ steps, and if this doesn't halt yet, then $N$ halts, but if $M$ does halt, then $N$ counts up to $|y|^3$ and then halts. That is, if $M$ doesn't halt on $z$, then $N$ halts quickly, but if $M$ does halt on $z$, then $N$ takes a long time on large input.
Thus, $M$ fails to halt on $z$ if and only if $N$ runs in $100n^2+200$ time on all input of size $n$. So if we could decide the latter property, then we could decide the halting problem, which is impossible.
This argument actually shows more, namely, that the complement of the halting problem is $1$-reducible to the running-in-time-$f$ problem.
(A similar argument could make use of Yuval's suggestion, using the halting-on-empty-string problem instead of the halting problem, which would eliminate the need to consider nontrivial $z$.)