Five digit numbers divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetetion. Total number of such numbers are 

The given digits are $0,1,2,3,4,5$

Case (i) : FIve digit numbers excluding zero.

All five digit numbers formed by $1,2,3,4,5$ are divisible by 3, becasue sum of the digits of $1,2,3,4,5$ is 15, which is divisible by 3

The number of five digit numbers formed by $1,2,3,4,5$ is equal to the number of ways of arranging the five numbers among them selves, this can be done in $5!$ ways

Case (II) : Five digit numbers including zero. 

If zero included, the five digit numbers which are divisible by 3 must contain $0,1,2,4,5$because their sum is 12, which is divisible by 3

In this case of five digit numbers, first place the first digit any one of $1,2,4,5$, this can be done in four ways. 

The remianing four blanks can be filled with four digits in $4!$ ways.  

Hence, the number of five digit numbers which are divisible by 3 are $4·4!$

Therefore, the number of five digit numbers divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetetion is $5!+4\left(4!\right)=120+96=\boxed{216}$