Bohr postulated that the angular momentum of such circular orbit would be quantized in multiples of $\hbar$, that is
$$L=n\hbar$$
This is a postulate, which means it's a happy idea that we consider true, but he gavee no proof.
Now, if we apply a classical definition, $\vec{L}=m \vec{r}\times \vec{v}$, and, for a circular orbit, we have
$$L=mrv$$
So the equation is $mrv=n\hbar$. Now, if we consider that the only attraction is the electrostatic one, we apply Newton's 2nd law. The electrostatic force must act as the centripetal force, so
$$\dfrac{1}{4\pi\epsilon_0}\frac{q_e^2}{r^2} = m \frac{v^2}{r}$$
$$\dfrac{1}{4\pi\epsilon_0}\frac{q_e^2}{mr} = v^2$$
Then replace
$$ m r v = n \hbar$$
$$ m^2 r^2 v^2 = n^2 \hbar^2$$
$$ m^2 r^2 \dfrac{1}{4\pi\epsilon_0}\frac{q_e^2}{mr} = n^2 \hbar^2$$
$$ m r \dfrac{1}{4\pi\epsilon_0}\frac{q_e^2}{1} = n^2 \hbar^2$$
$$ r = n^2 4\pi\epsilon_0 \dfrac{\hbar^2}{mq_e^2} = n^2 \cdot a_0 $$
with $a_0$ being $4\pi\epsilon_0 \dfrac{\hbar^2}{mq_e^2}$
Edit:
As I said, Bohr does not explain why it is like that, it is a postulate.
Then, Louis de Broglie came up with an explanation. The electron is like a stationary wave around the orbit.
The wave must be stationary (otherwise it would emit EM radiation). Such a stationary wave requieres a junction of the begining and the end. It's liek a usual stationary wave in which you join both extremes forming a circumference.
So, the junction of both endings requires the circumference lenght to be a multiple of the wavelenght
$lenght = n\cdot \lambda$, or $2\pi r= n \lambda $
Then include de Broglie's hypothesis and you have
$$ 2\pi r = n \frac{h}{mv} $$
which is the same condition from the Bohr's postulate. This was a curious explanation.
However, the real truth behind this solution is that it is solution to the Schrödinger's equation.