1. Identify the problem: We need to find the smallest prime number that cannot be written in the form $\left| 2^a - 3^b \right|$ with non-negative integers $a$ and $b$.

2. Check smaller primes: We start by verifying if smaller primes can be written in the form $\left| 2^a - 3^b \right|$.
$$\begin{aligned} &2 = 2^1 - 3^1, \\ &3 = 2^2 - 3^0, \\ &5 = 2^3 - 3^1, \\ &7 = 2^4 - 3^2, \\ &11 = 3^3 - 2^4, \\ &13 = 2^4 - 3^1, \\ &17 = 2^6 - 3^4, \\ &19 = 2^3 - 3^3, \\ &23 = 2^5 - 3^2, \\ &29 = 2^5 - 3^1, \\ &31 = 2^5 - 3^0, \\ &37 = 2^6 - 3^3. \end{aligned}$$
All primes less than 41 can be written in the form $\left| 2^a - 3^b \right|$.

3. **Check if 41 can be written in the form $\left| 2^a - 3^b \right|$:
- Case 1**: $2^a - 3^b = 41$
$$2^a \equiv 41 + 3^b \pmod{8}$$
Since $2^a \equiv 0 \pmod{8}$ for $a \geq 3$, we have:
$$41 + 3^b \equiv 0 \pmod{8} \implies 3^b \equiv -1 \pmod{8}$$
However, $3^b \pmod{8}$ cycles through $1, 3, 1, 3, \ldots$, so $3^b \equiv -1 \pmod{8}$ has no solutions.

- Case 2: $2^a - 3^b = -41$
$$2^a \equiv -41 + 3^b \pmod{8}$$
Since $2^a \equiv 0 \pmod{8}$ for $a \geq 3$, we have:
$$-41 + 3^b \equiv 0 \pmod{8} \implies 3^b \equiv 1 \pmod{8}$$
This implies $b$ must be even. Let $b = 2l$:
$$2^a = 3^{2l} + 41$$
Taking modulo 3:
$$2^a \equiv 1 \pmod{3}$$
This implies $a$ must be even. Let $a = 2k$:
$$2^{2k} = 3^{2l} + 41$$
Factoring:
$$(2^k - 3^l)(2^k + 3^l) = 41$$
Since 41 is prime, the only factor pairs are $(1, 41)$ and $(41, 1)$. This implies:
$$2^k - 3^l = 1 \quad \text{and} \quad 2^k + 3^l = 41$$
Solving these:
$$2^k = 21 \quad \text{and} \quad 3^l = 20$$
Neither of these are powers of 2 or 3, so no solutions exist.

Hence, we have exhausted all cases. Thus, the answer is $\boxed{41}$.