What is $DTIME(n^a)^{DTIME(n^b)}$? This might be embarrassing, but it turned out I don't know what is $DTIME(n^a)^{DTIME(n^b)}$. It is between $DTIME(n^{ab})$ and $DTIME(n^{a(b+1)})$ but where?

Update: There are three possible ways to define oracle machines.

1, They have an oracle tape, which they can read (see wikipedia).

2, They have an oracle tape, from which they can pose questions (see Arora-Barak book).

3, Same as two, but after each question the tape is erased (I could swear I heard this too).

I had 2 in mind, as the most common (?) definition.
In case of 3 it is easy to prove that the answer is $DTIME(n^{ab})$ (see Kristoffer's answer).
In case of 1, I am not even sure that the bounds I claim hold, but maybe it is not hard to prove, I have no clue about this version.

Update: I have realized that the number of tapes changes the power, so it is best to forget this question...
It seems your upper bound can be improved as follows: On input $x$ of length $n$, let $y_1,\dots,y_m$ be the oracle queries. Note that $|y_1|+\dots+|y_m| \leq c_1n^a$. The time to simulate the oracle machine to answer query $y_i$ is $c_2|y_i|^b$. Note now that $c_2(|y_1|^b+\dots+|y_m|^b) \leq c_2 (|y_1|+\dots+|y_m|)^b \leq c_2(c_1n^a)b = c_1c_2n^{ab}$. Thus the time to simulate the computation without the oracle is $c_1n^a + c_1c_2n^{ab} = O(n^{ab})$, which matches your lower bound.