Let us assume that the source voltage is $E$, the back emf developed in the armature coil is $e$, the armature resistance is $R$ and the armature current is $i$.

The efficiency of the motor is

$\eta =\frac{ie}{iE}=\frac{e}{200}=0.4$

$\Rightarrow e=80 \mathrm{V}$

The armature current is

$i=\frac{E-e}{R}$

$\therefore 10=\frac{200-80}{R}$

$\therefore R=12 \mathrm{\Omega }$