One approach to solving this is to first calculate the current the will flow through all three resistors:
The total resistance is R1+R2+R3 = 5280
I = V/R = 30/5280 ≈ 5,682 mA
Then the drop across R2 = I*R2 = 5,7mA * 1000 ≈ 5.682 V.
So this is based on the fact that V=I*R and there is one current running in the entire chain, ie., it is the same through each resistor.