Identify the species

The ion is hydrogen-like lithium, i.e. $\mathrm{Li^{2+}}$ (one-electron system).

For a hydrogen-like atom, $E_n = -\frac{13.6 Z^2}{n^2}\ \text{eV}$ where $Z=3$ for lithium.

Write energies of the required levels

From first orbit to third orbit means transition from $n=1$ to $n=3$.

$E_1 = -13.6\times 3^2 = -13.6\times 9 = -122.4\ \text{eV}$

$E_3 = -\frac{13.6\times 9}{3^2} = -13.6\ \text{eV}$

Find the excitation energy

Required photon energy: $\Delta E = E_3 - E_1 = (-13.6)-(-122.4) = 108.8\ \text{eV}$

Use $E = \dfrac{hc}{\lambda}$

Given: $hc = 1242\ \text{eV nm}$

So, $\lambda = \frac{1242}{108.8}\ \text{nm}$

$\lambda \approx 11.42\ \text{nm}$

Convert into the required form

Since $1\ \text{nm} = 10^{-9}\ \text{m}$ therefore, $11.42\ \text{nm} = 11.42\times 10^{-9}\ \text{m} = 114.2\times 10^{-10}\ \text{m}$

Hence, $x \approx 114$

Compare with stored answer

Stored correct answer = $114$.

This matches the derived answer.