The correct set of 4 quantum numbers for the outermost elsectrons of rubidium (Z=3) is?

First, rubidium is not Z=3; Z=37. You have 4 quantum numbers to find: n,l,m, and s.

You need to know the electron configuration for rubidium, which is [Kr] 5s1. This tells us that rubidium is on the 5th row of the periodic table, so n=5. Rubidium is in the s-block of the periodic table (period I), so l=0. l would be 1 if it were in the p-block, 2 in the d-block, and 3 in the f-block. Because l=0, m=0, since m must be an integer between -l and l inclusive. s depends on the orientation of the magnetic moment of the electron in space, but it has values of either +1/2 or -1/2. Convention states that the first half of electrons in a subshell are +.

Your quantum numbers are therefore:

n=5

l=0

m=0

s=1/2