What actually makes a Bohr's radius stable?

I know that this theory is now invalid and we have quantum model but I want to know why Bohr's model got attention in the first place.

This was the first model that made a successful prediction of atomic (hydrogen) emission frequencies (the Rydberg formula). Other than that, Bohr's model doesn't give too much insight into how the atoms work. It just consisted of "strange" postulates, created in order to avoid collapse of the atom.

Just 12 years after its introduction, Heisenberg and Schrödinger made the first steps to create the actual consistent theory that's now called quantum mechanics. It's just a historical curiosity that Bohr's model is still being taught.

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 $4\pi\epsilon_0 \dfrac{\hbar^2}{mq_e^2}$

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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.

In classical physics an electron orbiting the nucleus will constantly lose energy by emitting electromagnetic radiation and eventually will collapse to the nucleus. But If you accept the fact, as Bohr did, that angular momentum is quantized then you can tell that a change in orbital radius cannot happen in small steps. Therefore the electron has to lose or gain a lot of energy for this change to happen. This is the stability of Bohr's model.