Is there a quick yet semi rigorous way to derive the Bohr radius of hydrogen?

Have a look at The Feynman Lectures on Physics, Volume 3 (chapter 2, section 2-4). There Feynman determines $a$ by minimizing the energy of the atom. The total energy is roughly $$ E = \frac{\hbar^2}{2ma^2} - \frac{e^2}{a}$$ Above you need to understand why equating the momentum $p$ with $\Delta\,p\approx \frac{\hbar}{a}$ is reasonable. Also, as defined in Volume I, $e^2$ here is the charge of an electron squared, divided by $4\pi\varepsilon_0$.

To minimize $E$ one differentiates wrt $a$ and set the result equal to $0$ and solve for $a$. $$\frac{dE}{da} = -\frac{\hbar^2}{ma^3} + \frac{e^2}{a^2}$$ Setting $\frac{dE}{da} = 0$ gives $a = \frac{\hbar^2}{me^2} \approx 0.5\times10^{-10}$m.

So, the "quick yet semi rigorous way to derive the Bohr radius hydrogen'' you're looking for is an application of the uncertainty principle.

From the Virial theorem, $\langle T\rangle = -\frac{1}{2}\langle V\rangle \rightarrow \langle E \rangle = \langle T\rangle + \langle V\rangle = \frac{1}{2}\langle V \rangle$. Since the potential is given by $V(r) = -\frac{e^2}{4\pi \epsilon_0 r}$, this tells you that in an energy eigenstate,

$$ \frac{e^2}{8\pi\epsilon_0} \langle \frac{1}{r}\rangle = \frac{13.6 eV}{n^2}$$


$$\langle \frac{1}{r}\rangle = \frac{13.6 \text{ eV}}{n^2} \left(\frac{8\pi\epsilon_0}{e^2}\right)$$

Inverting this$^\dagger$ and setting $n=1$ yields the Bohr radius

$$r = \frac{e^2}{8\pi\epsilon_0} \frac{1}{13.6\text{ eV}}\approx 5.3 \times 10^{-11}\text{ m}$$

$^\dagger$The Bohr radius provides the characteristic length scale for the problem. If you wish, you could recall that ground state wavefunction is of the form $\psi(r,\theta,\phi)=c e^{-r/a}$, and evaluate the expected value of $\frac{1}{r}$; if you do so, you will find $\langle \frac{1}{r}\rangle = \frac{1}{a}$, justifying the inversion above.

Alternatively, you could go back to the Bohr model, in which one has that the centripetal force is provided by the electrostatic force

$$\frac{mv^2}{r} = \frac{e^2}{4\pi\epsilon_0 r^2} \implies r = \frac{e^2}{4\pi \epsilon_0 mv^2}$$

and that the angular momentum is an integer multiple of $\hbar$: $$L = mvr = n\hbar \implies v^2 = \frac{n^2 \hbar^2}{m^2r^2}$$

which together yield that $$r = \frac{e^2}{4\pi\epsilon_0 m}\frac{m^2 r^2}{n^2 \hbar^2} \implies r = \frac{4\pi \epsilon_0n^2\hbar^2}{m e^2}$$

setting $n=1$ yields the Bohr radius

$$r = \frac{4\pi \epsilon_0\hbar^2}{m e^2} \approx 5.3 \times 10^{-11}\text{ m}$$

Assume from the Lyman series that $E_1 = -13.6$eV. Assume the electron is trapped in an infinite box, $$ E_1 = \frac{\pi^2 \hbar^2}{2m a_o^2} = 13.6 \, \mathrm{eV}$$ then $$a^2_o \approx 2.74 \times 10^{-20}\mathrm{m}^2$$ which means $a_o \approx 1.6 \times 10^{-10}$m.