Answer:
Intermediate.
Step-by-step explanation:
Hello,
In this case, we can rewrite the steps as:
![A + B \rightarrow C\ \ (fast)\\\\C + B \rightarrow D + E\ \ (slow)\\\\E + B \rightarrow F \ \ (very fast)](https://img.qammunity.org/2021/formulas/chemistry/college/9qkdxj38tlz9rhi6t4b3q6wiyyxfgfjq3f.png)
Thus, we can notice that in the fast step, C is present as a product but after that is consumed in the slow step, for that reason, and by cause of its formation-consumption behavior, it is properly described as an intermediate as it is not neither a starting-up substance (reactant in the first step) nor a final substance (product in the final step).
Best regards.