Given is a simple graph with $239$ vertices, such that it is not bipartite and each vertex has degree at least $3$. Find the smallest $k$, such that each odd cycle has length at most $k$.
1. **Understanding the Problem:**
   We are given a simple graph with 239 vertices, where each vertex has a degree of at least 3, and the graph is not bipartite. We need to find the smallest \( k \) such that every odd cycle in the graph has a length of at most \( k \).

2. **Constructing an Example:**
   To find such a \( k \), we can construct an example graph that meets the given conditions and check the lengths of its odd cycles.

3. **Constructing the Graph:**
   Consider the following construction:
   - Take 59 isolated \( K_4 \) graphs. Each \( K_4 \) is a complete graph on 4 vertices.
   - This gives us \( 59 \times 4 = 236 \) vertices.
   - Number the remaining 3 vertices as \( 1, 2, 3 \).
   - Connect vertices \( 1, 2, 3 \) to form a \( K_3 \) (a complete graph on 3 vertices).

4. **Connecting the \( K_4 \) Graphs:**
   - Connect vertex \( 1 \) to a single vertex in the first \( K_4 \).
   - Connect vertex \( 2 \) to a single vertex in the second \( K_4 \).
   - Connect vertex \( 3 \) to a single vertex in the third \( K_4 \).

5. **Verifying the Degree Condition:**
   - In each \( K_4 \), every vertex has a degree of 3.
   - The vertices \( 1, 2, 3 \) each have a degree of 3 (2 from the \( K_3 \) and 1 from the connection to a \( K_4 \)).

6. **Checking Odd Cycles:**
   - The \( K_3 \) formed by vertices \( 1, 2, 3 \) is a triangle, which is an odd cycle of length 3.
   - Each \( K_4 \) contains odd cycles of length 3 (triangles) and even cycles of length 4.
   - The connections between \( 1, 2, 3 \) and the \( K_4 \) graphs do not introduce any new odd cycles longer than 3.

7. **Conclusion:**
   - In this construction, every odd cycle has a length of 3.
   - Therefore, the smallest \( k \) such that each odd cycle has length at most \( k \) is \( k = 3 \).

The final answer is \( \boxed{ k = 3 } \).