Runtime complexity of a brute force factoring algorithm? (in terms of bits)
It is $(\lfloor \sqrt{N} \rfloor-1) \cdot O(n^2) = O(\sqrt{N} \log^2 N) = O(2^{\frac{n}{2}} n^2)$ in the worst case: for each of the $O(\sqrt{N})$ choices of the dividend you perform a division of two numbers of at most $n$ bits.

(When you factor $N$ you might repeat some of the divisions, but these can be safely ignored since there can be at most $n$ repeated divisions, for a total complexity of $O(n^3) = o(2^{\frac{n}{2}} n^2)$.)

Notice that this is the best bound we can get for your question under reasonable hypotheses.
Indeed, if the complexity of dividing two n-bit numbers is also $\Omega(n^2)$ and you apply the brute-force algorithm on a prime $N$:

$$
\sum_{i=2}^{\lfloor \sqrt{N} \rfloor} T(\log N, \log i)
\ge \sum_{i=\lceil \sqrt{N}/2 \rceil}^{\lfloor \sqrt{N} \rfloor} T\left(\frac{1}{2}\log N - 1, \frac{1}{2}\log N - 1\right) = \Omega(\sqrt N) \cdot \Omega(\log^2 N) = \Omega(\sqrt{N} \log^2 N),
$$

where $T(i,j)$ is the time it takes to divide a $i$-bit integer by a $j$-bit integer and I am assuming that $T(i,j)$ is non-decreasing w.r.t. $i$ and $j$.