To find the total number of five-digit numbers that are divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 without repetition, we need to consider a few key points:

1. The total number of digits available is 6 (0, 1, 2, 3, 4, 5).
2. The number should be a five-digit number, so the first digit cannot be 0.
3. The sum of the digits of a number divisible by 3 should also be divisible by 3.

Let's break down the solution into steps:

Step 1: Choose the first digit
Since the first digit cannot be 0, we have 5 options: 1, 2, 3, 4, 5.

Step 2: Choose the remaining four digits
We have 5 digits remaining (0, 1, 2, 3, 4, 5). However, we need to ensure that the sum of these four digits is divisible by 3.

To make the calculation easier, we can group the remaining digits into two sets:
Set A: {0, 3}, Set B: {1, 2, 4, 5}

We can choose the digits in two ways:
1. Choose all four digits from Set B: There are 4 digits in Set B, so the number of ways to choose all four digits from Set B is 4P4 = 4! = 4 * 3 * 2 * 1 = 24.
2. Choose three digits from Set B and one digit from Set A: There are 4 digits in Set B and 2 digits in Set A. The number of ways to choose 3 digits from Set B is 4P3 = 4! / (4-3)! = 4! / 1! = 4. The number of ways to choose 1 digit from Set A is 2P1 = 2! = 2. So, the total number of ways to choose three digits from Set B and one digit from Set A is 4 * 2 = 8.

Step 3: Calculate the total number of five-digit numbers
For each of the 5 options for the first digit, we have 24 ways to choose the remaining four digits from Set B and 8 ways to choose the remaining four digits (three from Set B and one from Set A). So, the total number of five-digit numbers is:
(5 * 24) + (5 * 8) = 120 + 40 = 160.

Therefore, the correct answer is option D) 216.