We are asked to determine the probability of landing on an odd number and then landing on a 6.
To do that we will use the product rule of probabilities:

Where:

To determine the value of the probability of A we need to have into account that there is only 1 odd number (7) out of 3 possible numbers, therefore, the probability is:

Now, to determine the value of the probability of "B" we need to have into account that there is only one number 6 out of 3 numbers therefore, we have:

Now, we substitute the values:

Now, we solve the operations:

Therefore, the probability is 1/9