(a) 216
A number is divisible by 3 when the sum of the digits of the number is divisible by 3.
Out of the given 6 digits, there are only two groups consisting of 5 digits whose sum is divisible by 3.
1+2+3+4+5 = 15
0+1+2+4+5 = 12
Using the digits 1, 2, 3, 4 and 5, the 5 digit numbers that can be formed = 5!
Similarly, using the digits 0, 1, 2, 4 and 5, the number that can be formed = 5!4! {since the first digit cannot be 0}
∴ Total numbers that are possible = 5! + 5! 4! = 240 24 = 216