Solution:

Answer: $324$.

We will count the number of possibilities for each digit in Bob's ID number, then multiply them to find the total number of possibilities for Bob's ID number.

There are $3$ possibilities for the first digit given any last $5$ digits, because the entire number must be divisible by $3$, so the sum of the digits must be divisible by $3$.

Because the first two digits are a number divisible by $2$, the second digit must be $2, 4, 6$, or $8$, which is $4$ possibilities.

Because the first five digits are a number divisible by $5$, the fifth digit must be a $5$.

Now, if the fourth digit is a $2$, then the last digit has two choices, $2, 8$, and the third digit has $5$ choices, $1, 3, 5, 7, 9$.

If the fourth digit is a $4$, then the last digit must be a $6$, and the third digit has $4$ choices, $2, 4, 6, 8$.

If the fourth digit is a $6$, then the last digit must be a $4$, and the third digit has $5$ choices, $1, 3, 5, 7, 9$.

If the fourth digit is an $8$, then the last digit has two choices, $2, 8$, and the third digit has $4$ choices, $2, 4, 6, 8$.

So there are a total of $3 \cdot 4 (2 \cdot 5 + 4 + 5 + 2 \cdot 4) = 3 \cdot 4 \cdot 27 = 324$ possibilities for Bob's ID number.