Missing question: what is the current drawn by the motor from the line?
Solution:
The potential difference between the two ends of the motor is
[tex]\Delta V= \epsilon + Ir[/tex]
where [tex]\epsilon[/tex] is the emf, I the current and r the internal resistance of the motor.
By re-arranging the equation and using the data of the problem, we can solve to find the current I:
[tex]I= \frac{\Delta V-\epsilon}{r}= \frac{120 V-105 V}{3.2 \Omega}=4.7 A [/tex]
