The De-Broglie wavelength of an electron in the $n=2$ state of a hydrogen atom can be calculated as follows:

Firstly, calculate the radius $r$ for the $n=2$ state using the Bohr radius formula:

$ r = 0.052 n^2 $

For $n=2$, the radius is:

$ \begin{aligned} r &= 0.052 \times 4 \\ &= 0.208 \, \text{nm} \end{aligned} $

We use the relationship $M v r = \frac{n h}{2 \pi}$, leading to the expression for the De-Broglie wavelength $\lambda$:

$ \lambda = \frac{h}{M v} = \pi r $

Substituting the calculated $r$ value:

$ \begin{aligned} \lambda &= \pi \times 0.208 \, \text{nm} \\ &= 3.14 \times 0.208 \, \text{nm} \\ &= 0.65317 \, \text{nm} \\ &\approx 0.67 \, \text{nm} \end{aligned} $

Thus, the De-Broglie wavelength of the electron in the $n=2$ state is approximately $0.67$ nm.