The radius of the second Bohr orbit for hydrogen atom is (planck’s constant (h) = $6.6262\times {10}^{-34}Js$ mass of electron = $9.1091\times {10}^{-31}Kg$charge of electron ; $\left(e\right)=1.6021\times {10}^{-19}C$permittivity of vacum $\left({\in }_0\right)=8.854185\times {10}^{-12}K{g}^{-1}{m}^3{A}^2$

Bohr radius  $\left(r_n\right)$  $={\in }_0{n}^2{h}^2$

$r_n=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}kZ}$

$k=\frac{1}{4\pi {\in }_0}$

$\therefore r_n=\frac{n^2{h}^2{\in }_0}{\pi m{e}^{2}Z}=n^2\frac{a_0}{Z}$

Where, m = mass of electron

                         e = charge of electron

                         h = Planck’s constant

                         k = Coulomb constant

$r_n=\frac{n^2\times 0.53}{Z}\AA$

Radius of  n^th  Bohr orbit for H – atom = 0.53 $n^2\AA$

$\therefore$Radius of 2^nd Bohr orbit for H – atom 

$=0.53\times {\left(2\right)}^2=2.12\AA$