Use Bohr model relation for radius

For hydrogen atom, $r_n = n^2 a_0$ where $a_0 = 5.3 \times 10^{-11}\,\text{m}$.

The given excited-state radius is $8.48 \times 10^{-10}\,\text{m}$ which corresponds to the third excited state, i.e. $n=4$.

Check: $r_4 = 4^2 a_0 = 16a_0 = 16(5.3\times 10^{-11}) = 8.48\times 10^{-10}\,\text{m}$ So this is consistent.

Use de Broglie standing-wave condition

For an electron in Bohr orbit, $2\pi r_n = n\lambda_n$ Therefore, $\lambda_n = \frac{2\pi r_n}{n}$

Since $r_n = n^2 a_0$, $\lambda_n = \frac{2\pi n^2 a_0}{n} = 2\pi n a_0$

Hence, $\lambda_n \propto n$

Find the ratio

Ground state: $n=1$

Third excited state: $n=4$

Therefore, $\frac{\lambda_1}{\lambda_4} = \frac{1}{4}$

Compare with options

The required ratio of de Broglie wavelengths of electron in ground state to that in the excited state is $\boxed{\frac{1}{4}}$

None of the given options $4, 3, 9, 16$ match this value.

Likely issue in the question/options

If the question intended the ratio of wavelengths in the excited state to the ground state, then $\frac{\lambda_4}{\lambda_1} = 4$ which matches option A.

So the stored answer appears to match the inverse ratio, not the ratio actually asked.