Why are electrons in the 3d orbital in a higher energy state although electrons in the 4s orbital are further out?

The #4s# orbitals are usually higher in energy than the #3d# (specifically, #"Sc"# through #"Kr"#), but some textbooks don't acknowledge or recognize that #"Ca"# is the border past which the orbital energies switch. This is discussed in detail here as well.

If you are interested, a detailed explanation for chromium's #3d^5 4s^1# configuration is here, and an argument that does NOT rely on a "half-filled subshell" sentiment can be found in the second half of this answer here.

Given that orbitals are only as high or low energy as they are large (as nucleus-electron interaction goes as #1//r#), orbitals that extend farther out are often (but not always) higher in energy.

It's a result of two competing factors, namely

1. quantum level #n#.
2. angular momentum #l#.

The #3d# has lower #n# by #1# (which lowers energy) but higher #l# by #2# (which raises energy), but the effect of lower #n# predominates for first-row transition metals compared to greater #l# (meaning the #3d# is lower in energy).

This can be seen in the radial density distribution functions, which show that the #4s# does have the farther radial extent in the qualitative picture based on the hydrogen atom only:

This was graphed in the Excel sheet linked here (see "Combined RDFs" tab) using known hydrogen atom wave functions if you want to explore it.

The #4s# vs. #3d# comparison in potential energy is borderline, but Miessler, Fischer, and Tarr have cited their potential energies here (Appendix B.9), and I have graphed them below:

Apparently, the #4s# has been higher in energy all the way from #"Sc"# to #"Zn"# (and through #"Kr"# as well, #Z = 21 - 36#), NOT the other way around as even Tro might say.

The given trend suggests an ambiguity in how #"Sc"# (#Z = 21#) is ionized as well, and that is gone over here.