No two marbles have the same color means, the final outcome of the three draws must be a permutation of
Blue, Green, Red
There are $3! = 6$ such permutations possible.
Now, probability of getting a Blue first, Green second and Red third $=\dfrac{10}{60}\times \dfrac{20}{60}\times \dfrac{30}{60}$
Required probability $=6\times \dfrac{10}{60}\times \dfrac{20}{60}\times \dfrac{30}{60} =\dfrac{1}{6}.$
Correct Answer: $B$