A three-digit number $x$ in base $10$ has a units-digit of $6$. When $x$ is written is base $9$, the second digit of the number is $4$, and the first and third digit are equal in value. Compute $x$ in base $10$.
1. Let \( x \) be a three-digit number in base 10, written as \( x = 100A + 10B + 6 \), where \( A \) and \( B \) are digits.
2. When \( x \) is written in base 9, it is given as \( x_9 = C4C \), where \( C \) is a digit in base 9.
3. Convert \( x_9 = C4C \) to base 10:
   \[
   x = C \cdot 9^2 + 4 \cdot 9 + C = 81C + 36 + C = 82C + 36
   \]
4. Equate the base 10 expressions:
   \[
   100A + 10B + 6 = 82C + 36
   \]
5. Simplify the equation:
   \[
   100A + 10B + 6 - 36 = 82C
   \]
   \[
   100A + 10B - 30 = 82C
   \]
6. Since \( x \) has a units digit of 6, we know that \( 100A + 10B + 6 \equiv 6 \pmod{10} \). This is consistent with the given form of \( x \).
7. To find \( C \), we need to solve for \( C \) such that \( 82C \) is a valid three-digit number and \( C \) is a digit in base 9 (i.e., \( 0 \leq C \leq 8 \)).
8. We also know that \( 100A + 10B + 6 \) must be a three-digit number, so \( 100 \leq 100A + 10B + 6 \leq 999 \).
9. From the equation \( 100A + 10B - 30 = 82C \), we can solve for \( C \):
   \[
   100A + 10B - 30 = 82C
   \]
   \[
   100A + 10B = 82C + 30
   \]
10. Since \( C \) must be a digit in base 9, we test possible values for \( C \):
    - For \( C = 5 \):
      \[
      82 \cdot 5 = 410
      \]
      \[
      100A + 10B = 410 + 30 = 440
      \]
      \[
      100A + 10B = 440
      \]
      \[
      A = 4, B = 4
      \]
      Thus, \( x = 100 \cdot 4 + 10 \cdot 4 + 6 = 400 + 40 + 6 = 446 \).

Therefore, the number \( x \) in base 10 is \( \boxed{446} \).