De-Broglie wavelength of an electron orbiting in the $n=2$ state of hydrogen atom is close to (Given Bohr radius $=0.052\text{nm}$):

According to Bohr's quantization condition, the circumference of the $n^{th}$ orbit is an integral multiple of the de-Broglie wavelength ($\lambda$):
$$2\pi r_n=n\lambda$$
The radius of the $n^{th}$ orbit is given by $r_n=a_0{n}^2$, where $a_0$ is the Bohr radius.For $n=2$:
$$r_2=a_0\times 2^2=4a_0$$
Substituting this into the wavelength equation:
$$2\pi (4a_0)=2\lambda ⟹\lambda =4\pi a_0$$
Given $a_0=0.052\text{nm}$:
$$\lambda =4\times 3.1415\times 0.052\approx 12.566\times 0.052\approx 0.653\text{nm}$$
The value is close to $0.67\text{nm}$.