Using only the digits $2,3$ and $9$, how many six-digit numbers can be formed which are divisible by $6$?
To determine how many six-digit numbers can be formed using only the digits $2, 3,$ and $9$ that are divisible by $6$, we need to consider the properties of numbers divisible by $6$. A number is divisible by $6$ if and only if it is divisible by both $2$ and $3$.

1. **Divisibility by $2$:**
   - A number is divisible by $2$ if its last digit is even. Since we are restricted to the digits $2, 3,$ and $9$, the only even digit available is $2$. Therefore, the last digit of our six-digit number must be $2$.

2. **Divisibility by $3$:**
   - A number is divisible by $3$ if the sum of its digits is divisible by $3$. Let’s denote our six-digit number as $abcde2$, where $a, b, c, d, e$ are the first five digits, and the last digit is $2$.

   - The sum of the digits of the number is $a + b + c + d + e + 2$. For this sum to be divisible by $3$, the sum $a + b + c + d + e$ must be congruent to $1 \pmod{3}$ (since $2 \equiv 2 \pmod{3}$).

3. **Possible values for $a, b, c, d, e$:**
   - The digits $3$ and $9$ are both $0 \pmod{3}$.
   - The digit $2$ is $2 \pmod{3}$.

   We need to find combinations of $a, b, c, d, e$ such that their sum is $1 \pmod{3}$. This can be achieved by having either $2$ or $5$ digits equal to $2$ among $a, b, c, d, e$.

4. **Case 1: Five $2$'s**
   - If all five digits are $2$, then the number is $222222$. This is a valid number and there is only $1$ way to form this number.

5. **Case 2: Two $2$'s and three other digits**
   - We need to choose $2$ positions out of $5$ for the digit $2$. The number of ways to choose $2$ positions out of $5$ is given by $\binom{5}{2}$.
   - The remaining $3$ positions can be filled with either $3$ or $9$. Each of these $3$ positions has $2$ choices (either $3$ or $9$).

   Therefore, the number of ways to arrange two $2$'s and three other digits is:
   \[
   \binom{5}{2} \times 2^3 = 10 \times 8 = 80
   \]

6. **Total number of valid six-digit numbers:**
   - Adding the number of ways from both cases, we get:
   \[
   1 + 80 = 81
   \]

Thus, the total number of six-digit numbers that can be formed using the digits $2, 3,$ and $9$ and are divisible by $6$ is $\boxed{81}$.