Indicate which is lower in energy in a many-electron atom:

3d or 4s
Would it be 4s [because it fills before 3d]?

That is correct is almost all cases. There is an exception for Cr and Cu; however, I think the intent of the problem clearly means the 4s (and for the reason you sited).

To determine which orbital is lower in energy in a many-electron atom, we need to consider the order in which orbitals are filled according to the Aufbau principle.

The Aufbau principle states that electrons will occupy the lowest energy orbitals available first before filling higher energy orbitals.

In the case of a many-electron atom, we can compare the energy levels of the 3d and 4s orbitals. The 3d orbital comes after the 4s orbital in the order of filling according to the Aufbau principle.

However, there are exceptions to this rule. Chromium (Cr) and Copper (Cu) are two examples where the 3d orbital is actually slightly lower in energy than the 4s orbital. This is due to electron-electron repulsion effects within the atom. In these cases, the electrons will preferentially fill the 3d orbital before completely filling the 4s orbital.

In most other cases, such as the one you mentioned, the 4s orbital is indeed lower in energy than the 3d orbital. As a result, the 4s orbital will be filled before the 3d orbital.

So to answer your question, yes, in most cases, the 4s orbital is lower in energy than the 3d orbital in a many-electron atom, and it fills before the 3d orbital.