If I understand this correctly, the total of the two dice is odd, and the number on the 20 sided die is a multiple of the number on the 6 sided die.
If the number on the 6 sided die is even, then the number on the 20 sided die must also be even since all multiples of an even number must be even. So any total in those cases would be even, and are excluded.
So just looking at rolling a 1, 3 or 5 on the 6 sided die, what is the probability for each that the 20 sided die will be a multiple, and will be even. The number on the 20 sided die must be even so the total will be odd.
For 1, any even number on the 20 sided die results in an odd total
P(1 on d6) x P(even on d20) = 1/6 x 1/2 = 1/12
For 3, the even multiples are 6. 12, and 18. Three numbers would satisfy the condition out of 20 possibilities, so that probability is 3/20.
P(3 on d6) x P(even multiple on d20) = 1/6 x 3/20 = 1/40
For 5, only 10 and 20 will result in an odd total so the probability of success on the d20 2/20 = 1/10
P(6 on d6) x P(10 or 20 on d20) = 1/6 x/1/10 = 1/60.
Since any of those subsets satisfy the conditions, and all of the are mutually exclusive, add the probabilities for the overall probability:
1/12 + 1/40 + 1/60 = 10/120 = 3/120 = 2/120 = 15/120 = 1/8, but that is not a choice, so is there more information?