Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We start with the relation between the energy of a photon and its wavelength:
$$
E = \frac{hc}{\lambda}
$$
Now, consider two closely spaced energy levels with a small energy difference $ \Delta E $. The corresponding wavelengths will also be slightly different, and we want to calculate the wavelength difference $ \Delta \lambda $.
For a small change in energy, we can express the change in wavelength by differentiating $ E = \dfrac{hc}{\lambda} $ with respect to $ \lambda $:
$$
dE = -\frac{hc}{\lambda^2} d\lambda
$$
Rearranging this equation to express $ d\lambda $ (which is $ \Delta \lambda $):
$$
\Delta \lambda = -\frac{\lambda^2}{hc} \Delta E
$$
Since we are only interested in the magnitude of the wavelength difference, we drop the negative sign:
$$
\boxed{\Delta \lambda = \frac{\lambda^2}{hc} \Delta E}\tag 1
$$
$$\color{blue}{\bf [b]}$$
We need to use the boxed formula above which we just derived.
The energy levels of the hydrogen atom (in eV) are given by the formula:
$$
E_n = \dfrac{-13.6 }{n^2}
$$
The energy difference between the two transitions is
$$
\Delta E = E_{21} - E_{20} = \frac{-13.6}{21^2} - \left[\frac{-13.6}{20^2} \right]
$$
$$
\Delta E = \frac{-13.6}{21^2} + \frac{13.6}{20^2} \tag 2
$$
Now we need to find $\lambda_{21\to 1}$ where we know that
$$
\lambda_{21\to 1} = \dfrac{hc}{\Delta E_{21\to1}}
$$
Plug the known;
$$
\lambda_{21\to 1} = \dfrac{(6.63\times 10^{-34})(3\times 10^8)}{ \left(\frac{-13.6}{21^2} + \frac{13.6}{1^2} \right)(1.6\times 10^{-19}) }$$
$$
\lambda_{21\to 1}=\bf 91.61 \;\rm nm\tag 3
$$
Plug (2) and (3) into (1);
$$
\Delta \lambda = \frac{(91.61 \times 10^{-9})^2}{(6.63\times 10^{-34})(3\times 10^8)}\left[ \frac{-13.6}{21^2} + \frac{13.6}{20^2}\right](1.6\times 10^{-19})
$$
$$\Delta \lambda=\color{red}{\bf 0.0213}\;\rm nm$$
