The decimal number $13^{101}$ is given. It is instead written as a ternary number. What are the two last digits of this ternary number?
To find the last two digits of \( 13^{101} \) in base 3, we need to compute \( 13^{101} \mod 3^2 \). Here are the detailed steps:

1. **Compute \( 13 \mod 9 \):**
   \[
   13 \equiv 4 \mod 9
   \]

2. **Use the property of exponents modulo \( 9 \):**
   \[
   13^{101} \equiv 4^{101} \mod 9
   \]

3. **Simplify the exponent using Euler's theorem:**
   Euler's theorem states that \( a^{\phi(n)} \equiv 1 \mod n \) for \( \gcd(a, n) = 1 \). Here, \( n = 9 \) and \( \phi(9) = 6 \) because \( \phi(9) = 9 \left(1 - \frac{1}{3}\right) = 6 \).

   Since \( 4 \) and \( 9 \) are coprime:
   \[
   4^6 \equiv 1 \mod 9
   \]

4. **Reduce the exponent modulo \( 6 \):**
   \[
   101 \mod 6 = 5
   \]

5. **Compute \( 4^5 \mod 9 \):**
   \[
   4^5 = 1024
   \]
   \[
   1024 \mod 9 = 7
   \]
   (since \( 1024 = 9 \times 113 + 7 \))

6. **Convert \( 7 \) from decimal to ternary:**
   \[
   7_{10} = 21_3
   \]

Thus, the last two digits of \( 13^{101} \) in base 3 are \( 21 \).

The final answer is \( \boxed{ 21 } \).