Find the smallest prime $p$ such that the digits of $p$ (in base 10) add up to a prime number greater than $10$.
1. We need to find the smallest prime number \( p \) such that the sum of its digits is a prime number greater than 10.
2. The smallest prime number greater than 10 is 11.
3. We need to find a prime number \( p \) such that the sum of its digits equals 11.
4. Let's list the two-digit numbers whose digits add up to 11:
   - \( 29 \) (2 + 9 = 11)
   - \( 38 \) (3 + 8 = 11)
   - \( 47 \) (4 + 7 = 11)
   - \( 56 \) (5 + 6 = 11)
   - \( 65 \) (6 + 5 = 11)
   - \( 74 \) (7 + 4 = 11)
   - \( 83 \) (8 + 3 = 11)
   - \( 92 \) (9 + 2 = 11)
5. We need to check which of these numbers are prime:
   - \( 29 \) is prime.
   - \( 38 \) is not prime (divisible by 2).
   - \( 47 \) is prime.
   - \( 56 \) is not prime (divisible by 2).
   - \( 65 \) is not prime (divisible by 5).
   - \( 74 \) is not prime (divisible by 2).
   - \( 83 \) is prime.
   - \( 92 \) is not prime (divisible by 2).
6. The smallest prime number from the list is \( 29 \).

The final answer is \( \boxed{29} \)