1. Use the energy formula for hydrogen-like ions

For a hydrogen-like ion, the energy of the electron in the $n$th orbit is

$E_n=-13.6\,\frac{Z^2}{n^2}\ \text{eV}$

where $Z$ is the atomic number.

For $\mathrm{Li}^{2+}$, we have:

$Z=3$

2. Identify the second excited state

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

So we need the energy required to remove the electron from $n=3$ to infinity.

3. Calculate the energy of the electron in $n=3$

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

4. Find the ionization energy from this state

To remove the electron completely, we must take it from $E_3$ to $0$ eV.

So required energy is

$0-(-13.6)=13.6\,\text{eV}$

5. Match with the given form

Given:

$13.6\,\text{eV}=x\times 10^{-1}\,\text{eV}$

Thus,

$x\times 10^{-1}=13.6$

$x=136$

6. Final answer

$\boxed{136}$