1. Ionisation energy of hydrogen in ground state

Given energy of electron in first shell of H atom: $E_1(\text{H})=-2.2\times10^{-18}\,\text{J}$

So, ionisation energy of H from ground state is: $I_H=+2.2\times10^{-18}\,\text{J}$

2. Ionisation energy of lithium atom

Lithium has configuration $1s^2\,2s^1$. The electron removed during first ionisation is the outermost $2s$ electron.

Using Bohr-like approximation for hydrogen-like outer electron: $E_n=-\frac{2.2\times10^{-18} Z^2}{n^2}$

For Li, outer electron is in second shell, so take $Z=3$ and $n=2$: $E=-2.2\times10^{-18}\times\frac{3^2}{2^2}$ $E=-2.2\times10^{-18}\times\frac{9}{4}$ $E=-4.95\times10^{-18}\,\text{J}$

Hence ionisation energy required is: $I_{\text{Li}}=4.95\times10^{-18}\,\text{J}$

3. Longest wavelength for ionisation

For threshold ionisation, $\frac{hc}{\lambda}=I_{\text{Li}}$

So, $\lambda=\frac{hc}{I_{\text{Li}}}$

Substitute values: $\lambda=\frac{(6.63\times10^{-34})(3\times10^8)}{4.95\times10^{-18}}$

$\lambda=\frac{19.89\times10^{-26}}{4.95\times10^{-18}}$

$\lambda\approx4.02\times10^{-8}\,\text{m}$

Thus, $\lambda=x\times10^{-8}\,\text{m}$ so $x\approx4.02$

Nearest integer: $\boxed{4}$

4. Comparison with stored answer

Stored correct answer = $4$.

Our derived answer also = $4$.