A) $\{ 2 , 9 , 16 \}$
2 is divisible by 1, 9 is not divisible by 2, 16 is not divisible by 3
B)$ \{ 3 , 5 , 15 \}$
3 is divisible by 1, 5 is not divisible by 2, 15 is divisible by 3
C) $\{ 2 , 6 , 24 \}$
2 is divisible by 1 , 6 is divisible by 2, 24 is divisible by 3
D) $\{ 4 , 15 , 30 \}$
4 is divisible by 1, 15 is not divisible by 2, 30 is divisible by 3
So, $\{2 , 6, 24 \}$ is totally ordered subset of $\{1 , 2, 3\}$
