The de-Broglie wavelength of an electron in the first Bohr orbit is :

$m v r_n=\frac{n h}{2 \pi} \Rightarrow p r_n=\frac{n h}{2 \pi} \Rightarrow \frac{h}{\lambda} \times r_n=\frac{n h}{2 \pi}$

$\Rightarrow \lambda=\frac{2 \pi r_n}{n}$, for first orbit n = 1 so $\lambda=2 \pi r_1$ = circumference of first orbit