Let the three digits be X (hundreds), Y (tens) and Z (units).
It's tens digit is an odd multiple of 3. The multiples of 3 are 3, 9, 12 etc., so Y (tens) has to be either 3 or 9 since it is a single digit.
The ones digit and tens digit are both divisible by 3 and the ones and hundreds by 2. So, the ones digit is divisible by both 2 and 3.
The only single digit number divisible by both 2 and 3 is 6. Therefore, Z = 6 (units).
It is also given that the hundred's digit is 2 less than the one's digit, i.e, it is 2 less than 6 = 4. Therefore, X = 4 (hundreds).
The two possibilities are are 436 and 496. Since 436 is not one of the 4 options, 496 is the answer.