Since Bohr's radius is given by

$$\begin{gathered} r=\frac{n^2{h}^2{\epsilon }_0}{{\mathrm{\pi me}}^2\mathrm{Z}} \\ r\propto \frac{1}{m} \\ \frac{r}{r_0}=\frac{m_e}{m_{\mu }} \\ r=\frac{r_0}{212}=\frac{0.53A}{212}=250fm \end{gathered}$$

Since $$\begin{gathered} F\propto m \\ E=E_0\left(\frac{m_{\mu }}{m_e}\right) \\ E=\left(13.6 eV\right)\left(212\right) \\ E=2883 eV \end{gathered}$$

Angular momentum is given as

$L=\frac{h}{2\mathrm{\pi }}$

For ground state $n=1$ so we have 

$L=\frac{h}{2\mathrm{\pi }}$