Total number of calls during insertion into binary tree
The key observation here is that the input is assumed to be such that all insertions fill level 0 (the root level), then completely fill level 1, then completely fill level 2, and so on.

In filling level $i$ we will have constructed $2^i$ new nodes, each of which will require $i+1$ calls to INSERT, so the total cost to fill level $i$ will be $(i+1)2^i$. Thus, the cost to fill levels $0, 1,\dotsc,  k-1$ will be
$$
\sum_{i=0}^{k-1}(i+1)2^i
$$
Now let's say we're finishing by filling level $k$ partially or fully. We're then inserting elements $n=2^k,2^k+1,2^k+2, \dotsc, 2^{k+1}-1$. Each of those will cost $k+1$ calls to INSERT, so for $n$ elements we will have used $n-2^k+1$ insertions, each of cost $k+1$ for a total of $(k+1)(n-2^k+1)$ calls. That must be added to the cost of filling the upper rows, so you get a total cost of inserting $n$ elements level-by-level equal to
$$
\left(\sum_{i=0}^{k-1}(i+1)2^i\right) + (k+1)(n-2^k+1)
$$