A prime number$ q $is called[b][i] 'Kowai' [/i][/b]number if $ q = p^2 + 10$ where $q$,  $p$, $p^2-2$,  $p^2-8$,  $p^3+6$ are prime numbers. WE know that, at least one [b][i]'Kowai'[/i][/b] number can be found. Find the summation of all [b][i]'Kowai'[/i][/b] numbers.

To solve the problem, we need to find a prime number \( q \) such that \( q = p^2 + 10 \) and the numbers \( q \), \( p \), \( p^2 - 2 \), \( p^2 - 8 \), and \( p^3 + 6 \) are all prime numbers. We will use modular arithmetic to narrow down the possible values of \( p \).

1. **Consider the condition \( p^2 - 8 \) being prime:**
   - If \( p \equiv 1 \pmod{7} \) or \( p \equiv 6 \pmod{7} \), then \( p^2 \equiv 1 \pmod{7} \).
     \[
     p^2 - 8 \equiv 1 - 8 \equiv -7 \equiv 0 \pmod{7}
     \]
     This implies \( 7 \mid (p^2 - 8) \), and since \( p^2 - 8 \neq 7 \), \( p^2 - 8 \) cannot be prime.

2. **Consider the condition \( p^2 + 10 \) being prime:**
   - If \( p \equiv 2 \pmod{7} \) or \( p \equiv 5 \pmod{7} \), then \( p^2 \equiv 4 \pmod{7} \).
     \[
     p^2 + 10 \equiv 4 + 10 \equiv 14 \equiv 0 \pmod{7}
     \]
     This implies \( 7 \mid (p^2 + 10) \), and since \( p^2 + 10 \neq 7 \), \( p^2 + 10 \) cannot be prime.

3. **Consider the condition \( p^2 - 2 \) being prime:**
   - If \( p \equiv 3 \pmod{7} \) or \( p \equiv 4 \pmod{7} \), then \( p^2 \equiv 9 \equiv 2 \pmod{7} \).
     \[
     p^2 - 2 \equiv 2 - 2 \equiv 0 \pmod{7}
     \]
     This implies \( 7 \mid (p^2 - 2) \), and since \( p^2 - 2 = 7 \), we have \( p^2 = 9 \), so \( p = 3 \).

4. **Check the condition \( p^3 + 6 \) being prime:**
   - For \( p = 3 \):
     \[
     p^3 + 6 = 3^3 + 6 = 27 + 6 = 33
     \]
     Since 33 is not a prime number, \( p \neq 3 \).

5. **Consider \( p = 7 \):**
   - Check if \( p = 7 \) satisfies all conditions:
     \[
     q = p^2 + 10 = 7^2 + 10 = 49 + 10 = 59
     \]
     - \( p^2 - 2 = 7^2 - 2 = 49 - 2 = 47 \) (prime)
     - \( p^2 - 8 = 7^2 - 8 = 49 - 8 = 41 \) (prime)
     - \( p^3 + 6 = 7^3 + 6 = 343 + 6 = 349 \) (prime)

Since all conditions are satisfied for \( p = 7 \), the only ***'Kowai'*** number is \( q = 59 \).

The final answer is \( \boxed{59} \).