Answer
a) ${\bf 4.26\times 10^{-14}}\;\rm m$, ${\bf 1.31 \times 10^7 }\;\rm m/s$
b) ${\bf 0.0164}\;\rm nm$
c) $\bf X-ray$
d) ${\bf 7.34\times 10^{13}}\;\rm rev $
Work Step by Step
$$\color{blue}{\bf [a]}$$
We can say that the muon behaves like an electron in a hydrogen-like ion with a charge $Z$, but the muon's differs here.
The mass of the is given by
$$m_\mu = {\bf 207} m_e\tag 1$$
And since the mass changed, the Bohr radius changed too.
Hence,
$$(a_B)_\mu=\dfrac{4\pi\epsilon_0\hbar^2}{m_\mu e^2}\times \dfrac{m_e}{m_e}$$
$$(a_B)_\mu=\overbrace{\dfrac{4\pi\epsilon_0\hbar^2}{m_e e^2}}^{a_B}\times \dfrac{m_e}{m_\mu}$$
$$(a_B)_\mu= \dfrac{m_e}{m_\mu}a_B$$
Plug from (1);
$$(a_B)_\mu= \dfrac{1}{207}a_B\tag 2$$
The ground state radius of the muon in the carbon is given by
$$r_1=\dfrac{(a_B)_\mu}{Z}$$
Plug from (2);
$$r_1=\dfrac{a_B}{207\;Z}$$
Plug the known;
$$r_1=\dfrac{0.0529\times 10^{-9}}{207(6)}=\color{red}{\bf 4.26\times 10^{-14}}\;\rm m$$
Now we need to find the speed on the ground state,
$$v_1=\dfrac{\hbar}{m_\mu r_1}=\dfrac{\hbar}{207\;m_e r_1}$$
Plug the known;
$$v_1=\dfrac{1.055\times 10^{-34}}{207(9.11\times 10^{-31})(4.26\times 10^{-14})}$$
$$v_1 =\color{red}{\bf 1.31 \times 10^7 }\;\rm m/s$$
$$\color{blue}{\bf [b]}$$
First, we need to find the energy levels of hydrogen-like atom for this case.
$$ E_n = -Z^2 \dfrac{(E_1)_\mu}{n^2}\tag 3$$
where
$$(E_1)_\mu=\dfrac{1}{(4\pi\epsilon_0 )}\dfrac{e^2}{2(a_B)_\mu}\times \dfrac{a_B}{a_B}$$
$$(E_1)_\mu=\overbrace{\dfrac{1}{(4\pi\epsilon_0 )}\dfrac{e^2}{2a_B}}^{=13.6\;\rm eV}\times \overbrace{\dfrac{a_B}{(a_B)_\mu}}^{=207}$$
$$(E_1)_\mu=(207)(13.6)=\bf 2815\;\rm eV $$
Plug into (3);
$$ E_n = -Z^2 \dfrac{2815\;\rm eV}{n^2} \tag 4$$
Now we need to find $E_{\rm photon}=E_{2\rightarrow1}$ for the muon transition.
$$E_{2\rightarrow1}=E_{\rm photon}=|E_1-E_2|$$
Using (4);
$$E_{\rm photon}=\left| -\left(6^2\right) \dfrac{2815 }{1^2} -\left[ -\left(6^2\right) \dfrac{2815 }{2^2} \right]\right|=\bf 76005 \;\rm eV$$
Hence, its wavelength is given by
$$\lambda_{2\rightarrow1}=\dfrac{hc}{E_{\rm photon}}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(76005 \times 1.6\times 10^{-19})}$$
$$\lambda_{2\rightarrow1}=\color{red}{\bf 0.0164}\;\rm nm$$
$$\color{blue}{\bf [c]}$$
We know that the range of wavelength regions is given by:
Infrared (IR): $700 \; \rm{nm}$ to $1 \; \rm{mm}$
Visible light: $400 \;\rm{nm}$ to $700 \;\rm{nm}$
Ultraviolet (UV):$10 \; \rm{nm}$ to $400 \; \rm{nm}$
X-rays:$0.01 \; \rm{nm}$ to $ 10 \; \rm{nm}$
And since $0.0164 \; \rm{nm}$ is much shorter than even ultraviolet light, it falls into the $\bf X-ray$ part of the spectrum.
$$\color{blue}{\bf [d]}$$
The number of revolutions made by a muon before it dies is given by
$$N=\dfrac{\Delta t}{T}\tag 5$$
where $\Delta t=1.5\;\mu\rm s $ is the lifetime of the muon and $T$ is the orbital period of the muon which is given by
$$T =\dfrac{2\pi r_1}{v_1}$$
Plug into (5);
$$N=\dfrac{v_1\Delta t}{2\pi r_1} $$
Plug the known
$$N=\dfrac{(1.31 \times 10^7)(1.5\times 10^{-6})}{2\pi (4.26\times 10^{-14})} = \color{red}{\bf 7.34\times 10^{13}}\;\rm rev $$
