The magnitude of the orbital angular momentum L of an electron in a certain atom is equal to 3.464`barh` . Which one of the following numbers could...

I don't think your friend's method is entirely sound, but 4 is the correct answer. Here's another way of doing it. 

We know the relationship between the principal quantum number, n, and the angular quantum number l, is l = n-1. So if we find l, we just add one to find n. 

We also know that the orbital angular momentum is related to the angular quantum number through the equation L = sqrt(l(l+1)) hbar

We're given L = 3.464hbar, so the equation can be rearranged pretty easily.

3.464hbar = sqrt(l(l+1)) hbar....cancel the hbars

3.464 = sqrt(l(l+1))...square both sides

12 = l(l+1)...expand

12 = le2 + l...rearrange and solve quadratically

`l^2 + l - 12 = 0` 

`(l+4)(l-3)` ...so l can be -4 or 3. We only want positive numbers, so l is 3, and n is 4.