Answer: B

Small Hint:

With at least $6$ faces each, a sum of $7$ always has $6$ ways, so a sum of $10$ has $8$ ways

Big Hint:

Let the dice have $a\le b$ faces; count outcomes for sums $10$ and $12$ and use $\text{ways}(12)=\tfrac{ab}{12}$

Solution:

Let the dice have $a\le b$ faces. Since both have at least $6$ faces, a sum of $7$ occurs in exactly $6$ ways, so a sum of $10$ occurs in $6\div\tfrac34=8$ ways.

The number of ways to roll $10$ is $\min(a,9)$ $-\max(1,10-b)+1=8.$ A sum of $12$ has probability $\tfrac{1}{12},$ so it occurs in $\tfrac{ab}{12}$ ways.

Having $8$ outcomes for a sum of $10$ requires $a\ge8$ and $b\ge9,$ so necessarily $a+b\ge17.$ For $(a,b)=(8,9),$ a sum of $10$ has $8$ outcomes, while a sum of $12$ has the $6$ outcomes with the first die showing $3,4,\ldots,8.$ Since $\frac{6}{8\cdot9}=\frac{1}{12},$ both conditions hold and the lower bound $17$ is attained.

Thus, the correct answer is B.