The magnitude of the orbital angular momentum L of an electron in a certain atom is equal to 3.464barh . Which one of the following numbers could be the principal quantum number n of the electron? My buddy says it equals 4 because 3.464hbar= 3.64*10^-34 or really 3.64 rounded to 4. I don't believe that is the proper method. The answer key does say 4.

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.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qangm.html