Step 1: Start with Bohr's rule
$mvr = \dfrac{nh}{2\pi}$.
Step 2: Use $\lambda = \dfrac{h}{mv}$
Then $mv = \dfrac{h}{\lambda}$, so $\dfrac{h}{\lambda}r = \dfrac{nh}{2\pi}$.
Step 3: Simplify
$\lambda = \dfrac{2\pi r}{n}$.
Step 4: Check
For $n = 1$, one wavelength fits the circumference, as for a standing wave.
Step 5: Link to Bohr quantisation
The condition $2\pi r = n\lambda$ shows that the allowed orbits are those where a whole number of de Broglie waves fits around the circle. For $n = 2$ the circumference holds exactly two wavelengths. The options with $n\pi r$ or $\frac{nr}{\pi}$ give wavelengths that do not scale properly with the circumference.
Final Answer:
The wavelength is 2 pi r / n. This is option (C).
\[ \boxed{\text{(C) }\frac{2\pi r}{n}} \]