**Concept: The question tests on the area of Combinatorics**

**Solution:**

To be divisible by 3, the sum of the digits should be divisible by 3.

(a) We can have 0 + 1 + 2 + 4 + 5 = 12 which is divisible by 3.

The 1st place cannot have 0. So 4 of these numbers (1, 2, 4, 5) can come in the 1st place

the 2nd place can now have any 4 numbers (3 remaining numbers from above and 0)

The total numbers possible with these digits = 4 * 4 * 3 * 2 * 1 = 96

(b) We can cave 1 + 2 + 3 + 4 + 5 = 15

The total numbers possible here is 5! = 120

Therefore the total numbers possible = 96 + 120 = 216

(**option d)**