Meaning of the first dip in Franck–Hertz experiment
The first dip at $10.2\,\text{V}$ means an electron must gain energy $E = eV = 10.2\,\text{eV}$ to excite the hydrogen atom from ground state to its first excited state.

Energy of emitted photon
When the atom returns from the first excited state to the ground state, it emits a photon of the same energy: $E_{\gamma} = 10.2\,\text{eV}$

Use photon energy relation
$E = \frac{hc}{\lambda}$ Hence, $\lambda = \frac{hc}{E}$

Substitute given values
Given: $hc = 1245\,\text{eV nm}$ and $E = 10.2\,\text{eV}$

Therefore, $\lambda = \frac{1245}{10.2}\,\text{nm}$

Calculate
$\lambda \approx 122.06\,\text{nm}$

Final integer answer
$\boxed{122}$ nm