Since, the efficiency of an electric motor is given by
$\eta=\frac{\text { output power }}{\text { input power }}$
From the above relation, it is quite clear that maximum output power corresponds maximum efficiency of motor.
Now, output power is given by
$=e I=\frac{e(E-e)}{R}$ ... (i)
To obtain maximum output power differentiating E (i) with respect to $e$ which will be equal to zero.
So, $ \frac{d}{d e}\left[\frac{e(E-e)}{R}\right]=0$
$ \Rightarrow e=\frac{E}{2}$
Thus, when back emf becomes equal to half of the applied emf, the efficiency of motor will be maximum.