Answer: B

Solution:

If $N = 6\cdot(\text{digit sum}),$ then since the digit sum of a number below $1000$ is at most $27,$ we have $N \le 162.$

For a two-digit number $10t + u = 6(t + u)$ gives $4t = 5u,$ forcing $t = 5$ and $u = 4,$ so $N = 54.$ A one-digit number would need $6u = u,$ impossible for $u \gt 0.$ A three-digit number $100h + 10t + u = 6(h + t + u)$ gives $94h + 4t = 5u,$ whose left side is at least $94$ while the right side is at most $45,$ so there is no solution.

Hence exactly one number, $54,$ works.

Thus, the correct answer is B.