${\mathrm{Li}}^{2+}\to {\mathrm{Li}}^{3+}+e^{-}$. Energy required for this process. Given ionisation energy for ground state of hydrogen atom is $2.17\times {10}^{-18}\text{ J}$.

The correct option is:

$19.53\times {10}^{-18}\text{J}$

Step-by-step Explanation:

1. Identify the given system:
The given process represents the ionization of a lithium ion:
${\mathrm{Li}}^{2+}\to {\mathrm{Li}}^{3+}+e^{-}$
A neutral lithium atom has 3 electrons. Therefore, the lithium ion ${\mathrm{Li}}^{2+}$ contains only 1 electron. Since it contains a single electron, it is a hydrogen-like species, allowing us to use Bohr's model of the atom to calculate the energy required for its ionization.

2. Formula for the Ionization Energy of a hydrogen-like species:
The energy required to remove an electron from the ground state of a hydrogen-like atom or ion is given by the relation:
$E_{\text{ion}}=E_{\text{H}}\times Z^2$
where:
- $E_{\text{H}}$ is the ionization energy of a hydrogen atom in its ground state ($2.17\times {10}^{-18}\text{J}$).
- $Z$ is the atomic number of the element.

3. Substitute the values:
For Lithium ($\mathrm{Li}$), the atomic number is $Z=3$.
Substituting $Z=3$ and $E_{\text{H}}=2.17\times {10}^{-18}\text{J}$ into the equation gives:
$E_{\text{ion}}=2.17\times {10}^{-18}\text{J}\times 3^2$

4. Calculate the final energy value:
$E_{\text{ion}}=2.17\times {10}^{-18}\text{J}\times 9$
$E_{\text{ion}}=19.53\times {10}^{-18}\text{J}$