What is the voltage drop across the first light bulb?

v=6/3=2v                                     .

V(1) = V*1/(1+1+1) = 6*1/3 = 2 Volts 

You could use Ohm's law, use total resistance and voltage to figure the current, then figure that current against 1 ohm.

But you can also use common sense.  If there are three 1-ohm loads in series, the voltage is equally distributed between them.  Each resistor has 2 volts across it.  So the answer is D.