Hi,
Both the defensive ones are to be picked.This can be done in 1 way
From the remaining 8 one has to be picked in 8C1 ways.
Total number of ways of picked 3 out of 10 is 10C3
So, probability = 1*8C1/10C3 = 8/120 = 1/15

Hence, A

Coming to your method.
1st defective is selected in 2/10 ways
After it is picked 2nd one is picked in 1/9 ways
Third one is picked in 8/8 ways
So, probability is 1/5*1/9*1 = 1/45
But, here we are considering the order. In this case, the non-defective one is picked 3rd. It can be picked first or second as well
When non-defective one is picked first, probability is (8/10)*(2/9)*(1/8) = 1/45
When non-defective one is picked second, probability is (2/10)*(8/9)*(1/8) = 1/45
So, probability is 3(1/45) = 1/15.

Answer is A.
The answer should be solved in this way:
You have in all 10C3 choices that is 120 choices, so this is your denominator.
In the Numerator you have 2 fixed choices to get the 2 defectives so 2C2 * 8C1(which is any assembly out of the remaining 8)
Hence, 8/120 = 0.0667