You are building a simple Series circuit loop where two bulbs, one resistor, and one switch are used. The resistance of each bulb is 3 Ohms and the resistance of the
resistor is 3 Ohms. You are using a common 9-V Alkaline battery. What is the current
through the circuit?
0.67 Amps
0.67 Volts
1 Amp
1 Volt

Ohm's law tells you that i = v/r where v is equivalent voltage and r is equivalent resistance of the circuit
since all circuits have the same voltage, your question is really asking which circuit has the lowest total resistance, since the lower the resistance, the higher the current
circuit a has a resistance of 5 ohms; circuit c is in series so has a resistance of 10 ohms; circuit b is in parallel and therefore has a resistance of 2.5 ohms (review your notes on adding resistances in parallel and series); therefore b has the highest current