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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.
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