Step 1: Plan:
Use the ratio of speeds, since $f$ is the same.
Step 2: Steps:
$\frac{\lambda_{\text{brass}}}{\lambda_{\text{air}}} = \frac{v_{\text{brass}}}{v_{\text{air}}} = \frac{3500}{350} = 10$. Faster speed means longer wavelength, so it increases $10$ times. A drop by $10$ or $20$ times would need a slower medium.
Final Answer:
The wavelength increases by a factor of $10$, option (C).
\[ \boxed{\text{Increases by a factor of 10}} \]