16937=16926+11,
16926 is completely divisible and 11 is remaining.
So, remainder of $\frac{11^{30}}{31}$ and $\frac{(16937)^{30}}{31}$ will be same.
$\frac{11^{30}}{31}$ = $\frac{(11^{6})^{5}}{31}$
$\frac{(11^{6})}{31}$ gives 4 as remainder , So $\frac{4^{5}}{31}$ is remaining which gives 1 as remainder.
Hence,Option(A)1.