A muon is an unstable elementary particle of mass nearly $200 m_e$ and charge $\pm e$.

Here, a negative muon is given bound to a proton. 

So, $m=200 {\mathrm{m}}_e$ and $M=1836 {\mathrm{m}}_e$

(as mass of a proton is $1836$ times of mass of electron)

So, reduced mass of system, 

$m^{\text{'}}=\frac{mM}{m+M}=\frac{200m_e\times 1836m_e}{200m_e+1836m_e}\approx 180m_e$

As mass of muon is comparable to mass of proton, we have to take account of motion of nucleus. That's way we are calculating reduced mass.

Now, as energy of an electron in $n\mathrm{th}$ orbit is of $\mathrm{H}$-atom,

$E_n=\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}$

So, energy of a muon of muonic atom is 

$E_n^{\text{'}}=\frac{m^{\text{'}}{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}\Rightarrow E_n^{\text{'}}=180E_n$

Now, considering a Balmer transition, $n=3$ to $n=2$.

$\therefore \mathrm{\Delta }{E}^{\text{'}}=\frac{hc}{\lambda }$

or $\mathrm{\Delta }{E}^{\text{'}}=E_{n=3}^{\text{'}}-E_{n=2}^{\text{'}}=\frac{hc}{\lambda }$

$\Rightarrow 180\left(E_{n=3}-E_{n=2}\right)=\frac{hc}{\lambda }$

$\Rightarrow 180\left(\frac{-13.6}{3^2}-\frac{-13.6}{2^2}\right)=\frac{hc}{\lambda }$

$\Rightarrow 180\times 189 \mathrm{eV}=\frac{1240\left(\mathrm{eV}·\mathrm{nm}\right)}{\lambda \left(\mathrm{nm}\right)}$

$\Rightarrow \lambda \left(\mathrm{nm}\right)=3.6 \mathrm{nm}$

This emitted radiation is $X$-rays.

Note $X$-rays have a  wavelength range of $0.01 \mathrm{nm}$ to $10 \mathrm{nm}$.