I have been able to figure out the resistance for this problem, but I don't know how to get the voltage drop. I missed this class and do not have any formula for the voltage drop.

You decide to magnetize a steel nail by winding an insulated wire around it and connecting the stripped ends to the terminals of a battery. To limit the current and avoid draining the battery too quickly, you connect a flashlight bulb in series with the battery and wire wrapped around the nail. When the current in the wire is 45mA, the voltage across the coiled wire is 0.25 volts and the voltage across the bulb is 3.12 volts.

What is the resistance of the bulb? 69.3 ohms
What is the combined resistance of the bulb and coil? coil is 5.56 ohms = 74.89
What the combined voltage drop
What is the power drop across the bulb?

To limit the current and avoid draining the battery too quickly, you connect a flashlight bulb in series with the battery and wire wrapped around the nail. When the current in the wire is 45mA, the voltage across the coiled wire is 0.25 volts and the voltage across the bulb is 3.12 volts.

What is the resistance of the bulb? 69.3 ohms correct

What is the combined resistance of the bulb and coil? coil is 5.56 ohms = 74.89 correct

What the combined voltage drop
3.12+.25 = 3.37 volts

What is the power drop across the bulb?
(more correctly "power in the bulb" or "power lost in the bulb")
P = EI = 3.12*45 = 140 mW