Bohr Radius Confusion
Bohr's theory of the atom precedes quantum theory but does describe the Balmer and other lines in atomic hydrogen.  I've outlined the calculation so that you can see where the Bohr radius comes from.

He  assumed that the electron orbits the nucleus in a circle with a radius r and is determined by the balance of centripetal acceleration and Coulomb attraction towards the nucleus. For an electron of mass $m_e$ and speed $v$ this is 
$$\frac{m_ev^2}{r}=\frac{e^2/(4\pi\epsilon _0)}{r^2}$$
 where e is the charge on the electron and $4\pi\epsilon_0$ puts us into SI units, ( $\epsilon_0$ is the permittivity of free space). 

The total energy is the sum of kinetic and potential parts $$E=\frac{m_ev^2}{2}-\frac{(e^2/4\pi\epsilon _0)}{r}$$
and using the first equation gives 
$$E=\frac{e^2/(4\pi\epsilon _0)}{2r}$$

Bohr had to make the assumption that electrons circling in an orbit do not radiate energy, which is contrary to classical electrodynamics, but he then needed to determine which orbits were allowed. He assumed that the angular momentum is in quantised units of Plancks constant h and a (principal) quantum number n. This gives
$$ m_evr = n\frac{h}{2\pi}$$

which when combined with the first equation gives

$$ r=a_0n^2$$

which is the equation you ask about and $a_0$ is conventionally the symbol for the Bohr radius and is 0.0529 nm.
It is calculated as
$$a_0 = \frac{\hbar^2}{e^2/(4\pi\epsilon_0)m_e}$$

where $\hbar = h/2\pi$.
Re-expressing the  energy using $a_0$ gives
$$E=-\frac{e^2/(4\pi\epsilon_0)}{2a_0}\frac{1}{n^2}$$
which is the famous Bohr equation and explains the energy levels in hydrogen atoms and so the transitions between them.