To form a 5-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 without repetition, we need to consider the divisibility rule for 3. According to the rule, a number is divisible by 3 if the sum of its digits is divisible by 3.

To find the total number of ways to form such a number, we can break it down into several cases.

Case 1: 0 is the first digit
If 0 is the first digit, the remaining 4 digits can be chosen from the set {1, 2, 3, 4, 5} without repetition. There are 4! = 24 ways to arrange these digits. Therefore, there are 24 numbers in this case.

Case 2: 0 is not the first digit
In this case, the first digit can be any of the remaining 5 digits {1, 2, 3, 4, 5}. The remaining 4 digits can be chosen from the set {0, 1, 2, 3, 4, 5} without repetition. There are 5 choices for the first digit and 5! = 120 ways to arrange the remaining digits. Therefore, there are 5 * 120 = 600 numbers in this case.

Total number of ways
The total number of ways to form the 5-digit number is the sum of the numbers in each case. Therefore, the total number of ways is 24 + 600 = 624.

However, we need to consider that some of these numbers may have repetitions. For example, if the digits chosen are {1, 2, 3, 4, 5}, the number 12345 can be arranged in 5! = 120 ways. But we only want to count each combination once.

To remove the repetitions, we divide the total number of ways by the number of ways each combination can be arranged. In this case, each combination can be arranged in 5! = 120 ways. Therefore, the total number of distinct 5-digit numbers divisible by 3 is 624/120 = 5.2.

Since we cannot have a fraction of a number, we round down to the nearest whole number. Therefore, the correct answer is option A, 216.