Ionisation energy of $$\text{He}^+$$ is $$19.6 \times 10^{-18}$$ Jatom$$^{-1}$$. The energy of the first stationary state ($$n = 1$$) of $$\text{Li}^{2+}$$ is

For every hydrogen-like species (one-electron ion) the Bohr energy expression is

$$E_n = -\frac{Z^{2}}{n^{2}}\;E_H$$

Here $$E_H$$ is the magnitude of the ground-state energy of the hydrogen atom (which equals the ionisation energy of hydrogen), $$Z$$ is the nuclear charge and $$n$$ is the principal quantum number.

Step 1 Extract $$E_H$$ from the given data
For $$\text{He}^+$$, $$Z = 2$$ and for the ground state $$n = 1$$. The ionisation energy given in the question is

$$|\;E_1(\text{He}^+)| = 19.6 \times 10^{-18}\;\text{J atom}^{-1}$$

But $$|\;E_1(\text{He}^+)| = \frac{Z^{2}}{n^{2}}\,E_H = 4\,E_H$$ (because $$Z=2,\;n=1$$).

Therefore

$$E_H = \frac{19.6 \times 10^{-18}}{4} = 4.9 \times 10^{-18}\;\text{J atom}^{-1}$$

Step 2 Energy of the first stationary state of $$\text{Li}^{2+}$$
For $$\text{Li}^{2+}$$, $$Z = 3$$ and in the first Bohr orbit $$n = 1$$. Applying the formula,

$$E_1(\text{Li}^{2+}) = -\frac{Z^{2}}{n^{2}}\,E_H = -9\,E_H$$

$$\Rightarrow\;E_1(\text{Li}^{2+}) = -9 \times 4.9 \times 10^{-18}\; \text{J}$$

$$E_1(\text{Li}^{2+}) = -44.1 \times 10^{-18}\;\text{J} = -4.41 \times 10^{-17}\;\text{J atom}^{-1}$$

Result: the energy of the first stationary state of $$\text{Li}^{2+}$$ is $$-4.41 \times 10^{-17}\;\text{J atom}^{-1}$$.

Hence, the correct choice is
Option B which is: $$-4.41 \times 10^{-17}\;\text{J atom}^{-1}$$