Identify the required energy level

• Ground state corresponds to $n=1$.
• First excited state corresponds to $n=2$.
• Second excited state corresponds to $n=3$.

So we need the energy difference between $n=3$ and $n=1$.

Use the energy formula for a hydrogen-like ion

For a hydrogen-like ion, $E_n=-\frac{13.6 Z^2}{n^2}\,\text{eV}$

Therefore, $E_1=-13.6Z^2$ $E_3=-\frac{13.6Z^2}{9}$

Compute the excitation energy from $n=1$ to $n=3$

$\Delta E=E_3-E_1$ $\Delta E=-\frac{13.6Z^2}{9}-(-13.6Z^2)$ $\Delta E=13.6Z^2\left(1-\frac{1}{9}\right)$ $\Delta E=13.6Z^2\cdot\frac{8}{9}$

Given that this is $108.8\,\text{eV}$, $13.6Z^2\cdot\frac{8}{9}=108.8$

Solve for $Z$

First, $13.6\times 8=108.8$

So, $\frac{108.8}{9}Z^2=108.8$

Dividing both sides by $108.8$, $\frac{Z^2}{9}=1$ $Z^2=9$ $Z=3$

Match with the options

$Z=3$

Hence the correct option is C.