For muonic hydrogen, a muon (having mass 207 times the mass of electron) orbits around proton (having mass 1836 times the mass of electron). Approximated orbital radius in ground state for such system is $A\times {10}^{-B}$ in m. Here, A is integer lesser than 10. Find  $B-2A.$

Here,  $m=207m_e$ and  $M=1836m_e$ So the reduced mass is,  $m^{\text{'}}=\frac{mM}{m+M}$
 $=\frac{\left(207m_e\right)\left(1836m_e\right)}{207m_e+1836m_e}=186m_e$
  The orbit radius corresponding to $n=1$ is $r_1=\frac{h^2{\epsilon }_0}{\pi m_e{e}^2}$    
  Where,  $r_1=a_0=5.29\times {10}^{-11}m$
  Hence, the radius  $r^{\text{'}}$ that correspond to the reduced mass $m^{\text{'}}$  is  $r_1^{\text{'}}=\left(\frac{m}{m^{\text{'}}}\right)r_1=\left(\frac{m}{186m_e}\right)a_0$
$=2.85\times {10}^{-13}\approx 3\times {10}^{-13}m\Rightarrow A=3andB=13\Rightarrow \frac{B}{A}=\frac{13}{3}=4.33$