For any number $\mathrm{Y}$ in binary, Calculating $\mathrm{Y} \bmod 2^n$ is equivalent to stripping off all but the $\mathrm{n}$ lowest-order (right-most) bits of number $\mathrm{Y}$.
So in other words, if $\mathrm{Y}$ is $8$-bits number then $\mathrm{Y}$ mod $4$ is the same as $\mathrm{Y} \& 00000011$ (where $\&$ means bitwise-and)
Note that this works exactly the same in base-$10: \text{Y} \mod 10$ gives you the last digit of the number $\mathrm{Y}$ in base-$10, \mathrm{Y} \mod 100$ gives you the last two digits, etc.
So, If $\mathrm{M}$ is $32$ -bits number then $\mathrm{M} \mod 16$ is the same as $\mathrm{Y} \& 0000000 \mathrm{~F}$ (where $\&$ means bitwise-and)
Option C is the correct approach. The bitwise AND operation with $\textsf{0000000Fh}$ extracts the lowest $4$ bits of $\mathrm{M}$, and then the result is multiplied by $3$ by adding it to itself twice.
The question initially had ambiguity. It should have mentioned “mod” instead of “division”. The question was taken from a standard resource (Q 19 here) and the ambiguity came with it. Now we have corrected it & made it unambiguous.
Option $B$ would be correct if it was division instead of mod.
So, answer has been changed to $B$ or $C.$
