What is the runtime of Mergesort if we switch to Insertion Sort at logarithmic depth?
You know that MergeSort has a complexity of $O(n\log n)$. Why? We can write a recurrence for the running time of MergeSort:
$$ T(n) = 2T(n/2) + Cn, \quad T(1) = D. $$
(Assume for simplicity that $n$ is a power of 2.) The solution is $T(n) = O(n\log n)$. Why? Since we can write
$$ T(n) = 2T(n/2) + Cn = 4T(n/4) + Cn + Cn = 8T(n/8) + Cn + Cn + Cn = \cdots = (\log n)Cn + Dn. $$
That is, at each internal level the total complexity is $Cn$, there are $\log n$ levels, and at the bottom the complexity is $Dn$. It is instructive to rewrite it in terms of $n = 2^k$:
$$ T(2^k) = 2^1T(2^{k-1}) + C2^k = 2^2T(2^{k-2}) + C2^k + C2^k = \cdots. $$
In your case, you want to stop the recursion at depth $k/2$, and instead of expanding $2^{k/2}T(2^{k/2})$ using the recursion, apply the $En^2$ insertion sort algorithm. What would you get? (that's a question for you)