Why the energy of a state increases with increasing azimuthal quantum number $ell$?

I am reading Griffiths’ introduction to quantum mechanics. But I am confused about the relation between the state energy and azimuthal quantum number $ell$, which is discussed in Chapter 5.2.2 where the periodic table is introduced.

In the book, it is said that for $rm He$ with the main quantum number $n$ as 2, the effect of electron repulsion is to favor the lowest value of $ell$ which equals to 0; It is because the expectation value of $r$ (I suppose $r$ is the radius of orbit of electron to be discussed) increases with increasing $ell$, which is to say that angular momentum tends to throw the electron outward. Within a given shell, therefore, the state with lowest energy is $ell=0$, and the energy increases with increasing $ell$.

However, I think that, with a larger $r$, the distance between outer and inner shells electrons will be larger, and hence the energy due to electron repulsion would be smaller. By this way, I think it means that with a larger $ell$, the state would have a lower energy, which is a contradictory to the statement in the book.

So I would like to know where I made the mistake. By the way, in the book, it also mentions the screening effect. But I didn’t see how screening effect could affect the interaction between electrons.