The voltage all the way around the loop ... between the terminals of the battery ... is 36 volts total. It looks like 12 of those volts are across one lamp, and 12 of them are across the other lamp. That leaves the last 12 volts unaccounted for, and only one more component in the loop (the resistor), so the remaining 12 volts is across the resistor.
So far, we don't have enough information to calculate any current in the loop, or the power dissipated by any component. To go any farther, we'll need to take your word for it ... the current in the loop is 0.75 Ampere. Fine !
In a resistor . . . Resistance = (voltage) / (current)
Resistance = (12 volts) / (0.75 Amp) = 16 ohms .
Power dissipated by anything = (voltage across it) x (current through it).
Power dissipated by the resistor OR either lamp =
(12 volts) x (0.75 Amp) = 9 watts .
That's not a lot for a light bulb, but it IS for a resistor. That baby has to be physically large (maybe like a lipstick), and cooled ... attached to a heatsink, or air blowing over it, or water flowing over it.
PS: I hope you're right about that 0.75 Amp.
