Step 1: Examine the form of the language.
The language $L$ is defined as the union of two simpler languages:
\[ L_1 = \{a^{2+3k} \mid k \ge 0\}, \quad L_2 = \{b^{10+12k} \mid k \ge 0\} \]
Each language consists of strings over a single symbol whose lengths follow a fixed arithmetic pattern.
Such languages are regular, and the union of two regular languages is also regular.
Step 2: Recall what the pumping lemma guarantees.
For any regular language, there exists a pumping length $p$ such that every string $w \in L$ with $|w| \ge p$ can be split as:
\[ w = xyz \]
with the conditions $|xy| \le p$, $|y| \ge 1$, and $xy^iz \in L$ for all $i \ge 0$.
The pumping length depends on the structure of the automaton recognizing the language.
Step 3: Choose a pumping length that works for both parts.
Strings in $L_1$ repeat in steps of $3$, while strings in $L_2$ repeat in steps of $12$.
To ensure that pumping works for any sufficiently long string from either subset, the pumping length must be large enough to cover both patterns.
Selecting a value that comfortably exceeds both cycle lengths ensures validity.
Step 4: Check the given choices.
(A) 3: Too small to handle strings from $L_2$.
(B) 5: Still smaller than the shortest string length in $L_2$.
(C) 9: Less than 10, so it fails for strings like $b^{10}$.
(D) 24: Large enough to accommodate both periodic structures. This choice works.
Step 5: Final conclusion.
A pumping length that is guaranteed to satisfy the pumping lemma for the language $L$ is:
\[ \boxed{24} \]