First, we determine what's the equivalent resistance of the 3Ω and 6Ω resistors
Using the quick method, it's
(3×6)÷(3+6) or 2Ω
So the circuit's effectively 3Ω in series with 2Ω, 5Ω in total.
Current according to Ohm's Law is
V = 5V, R = 5Ω so I = 1A So, Ohm's Law again to calculate the voltage drop across the 3//2
V = IR, I is 1, R is 2, so the answer's 2V.