"Given an algorithm, decide whether it runs in polynomial time" is this problem in NP?
No.  If the problem was polynomial-time verifiable, it would be solvable in exponential time, and thus decidable; but we already know that is not decidable.
Why in exponential time?  Because $V$ runs in time $|w|^k$, it can read at most $|w|^k$ bits of the input.  So, it suffices to enumerate all possible strings $c$ of length at most $|w|^k$, and run $V$ on each of them.  The running time will be about $2^{|w|^k}$, which is finite and thus enough to make the original problem decidable.