When two lamps are connected in series to a battery, the electrical resistance that the battery senses is?

More than the resistance of either lamp, because resistances in series are simply added together.

EQUAL TO BOTH LAMPS RESISTANCE ADDING TOGETHER .

second answer is correct provided, the right value of resistance is used. However lamp resistance does depend on voltage applied. If no voltage (or very low voltage) is applied, the resistance can be 1/12 th of its resistance at nominal voltage. Its resistance at half the nominal voltage is lesser than this.

The battery does not sense the electrical resistance. A battery can be characterized as a voltage source in series with a resistor. When a load with a lower resistance is connected to the battery the voltage at the battery is lower than with a higher resistance load. Using ohms law and treating the battery resistance in series with the load resistor as a voltage divider the voltage changes due to a different load resistor can be calculated.

Ohms law is E=IR, E is the EMF in Volts, I is the electrical current in Amps and R is the resistance in Ohms.

The voltage divider is described as a voltage source loaded by two resistors in series.  The voltage source's voltage is divided between the two resistors. The current flowing through both resistors is calculated by using Ohms law.  The voltage is the voltage source's voltage and the resistance is the sum of both resistors. The voltage of each resistor can be calculated using Ohms law also. The current is the same for both resistors and the resistance is the resistance of each resistor.  

If you don't know the resistance of the battery then you can load the battery with a resistor. If you place a resistor on the battery that lowers its voltage to 1/2 the open circuit load then the two resistances are the same. This test could be dangerous so it is better to load it with a higher value resistor. If you adjust the load resistor so the value drops to 90% of the open circuit voltage then the battery resistance is the load resistor divided by 9.

I hope this helps.