Why is the complexity of N=pq for RSA considered as exponential time?

I'm studying about RSA algorithm for 16-bits and noticed that the complexity of N=pq is considered as exponential time.

In the algorithm, p and q are two random and distinct strong primes. All the prime numbers, except 2, are odd numbers. So, when p is multiplied by q, we will get an odd composite number, and the composite number can be easily factored. So, why is N=pq called the integer factorization problem and is hard to solve?

Note computational complexity, if not otherwise specified, is a function of the size of the information content of the inputs. We can think of that size as the number of bits needed to specify the input. (Another usable definition for the information content size is the length of an input string, where each character in the string is an element of a fixed finite "alphabet" set. Counting bits is the case where the alphabet has two elements.)

So a straightforward algorithm for finding the two non-trivial factors of a positive integer $N$ by trying division by $2$ and each odd number up to $\sqrt{N}$ does have complexity $O(\sqrt{N}) \subset O(N)$, but $N$ is not the information content variable we want. The number of bits needed to represent the input in the usual way is about $B = \log_2 N$. So the complexity is $O(\sqrt{2^B}) = O(2^{B/2})$. This is why we consider this algorithm exponential time.