It is equivalent to the computational Diffie-Hellman problem; if you can one of the two problems, you can solve the other (with a polynomial number of queries to the oracle which solves the other).
If you can solve the Diffie-Hellman problem, you can solve your problem: this can be seen by first noting that, with a Diffie-Hellman solver, given $g^b$, you can compute $g^{b^{-1}}$; this is done by computing $g^{b^k}$ (which can be done with $O(\log k)$ queries to the DH oracle) with $k=q-1$ (where $q$ is the order of the group). Once you have that, you can do another query with $g^{b^{-1}}$ and $g^a$ to obtain $g^{ab^{-1}}$, and once you have that, you can then compute $g^{acb^{-1}}$ directly (as you know the value of $c$).
If you can solve your problem, then you can solve the computational Diffie-Hellman problem with two queries; given $g^a$ and $g^b$, you can first do a query with $g^1$, $g^b$ and $c=1$ (to compute $g^{b^{-1}}$), and then with $g^a$ and $g^{b^{-1}}$ and $c=1$ to compute $g^{ab}$