Answer: A

Solution:

Rerolling one die, keeping two dice that sum to $s,$ wins with probability $\tfrac16$ when $s \le 6$ and $0$ otherwise. Rerolling two dice, keeping a die of value $v,$ wins with probability equal to the number of ways two dice sum to $7 - v,$ over $36;$ this is largest when $v$ is smallest. For $v=1,2,3,$ these probabilities are $\tfrac5{36},\tfrac4{36},\tfrac3{36},$ all greater than the reroll-all probability $\tfrac{15}{216}=\tfrac{5}{72};$ for $v\ge4,$ rerolling all is better.

Rerolling exactly two dice is strictly best precisely when the two smallest dice sum to at least $7$ (so rerolling one cannot reach $7$) while the smallest die is $1, 2,$ or $3$ (so keeping it beats rerolling all three).

Sort the roll as $u\le v\le w.$ If $u=1,$ the only possibility is $(1,6,6),$ with $3$ orderings. If $u=2,$ the possibilities $(2,5,5),(2,5,6),(2,6,6)$ have $3+6+3=12$ orderings. If $u=3,$ choose $v,w$ with repetition from $\{4,5,6\};$ the six resulting triples have $3+6+6+3+6+3=27$ orderings. Thus there are $3+12+27=42$ qualifying ordered rolls out of $216,$ a probability of $\dfrac{42}{216} = \dfrac{7}{36}.$

Thus, A is the correct answer.