Answer
a) $4\;\rm eV$
b) $3\;\rm eV$
c) $3\;\rm eV$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the ionization energy is the energy required to completely remove an electron from the atom, taking it from the ground state to an energy level of 0 eV which means the electron is no longer bound to the atom.
So, from the given graph, the ionization energy is given by
$$|E_1|=0-(-4)=\color{red}{\bf4}\;\rm eV$$
$$\color{blue}{\bf [b]}$$
We know that the energy of the emitted photon is given by
$$E_{\rm emitted}=\dfrac{hc}{\lambda}=\dfrac{(6.626 \times 10^{-34} ) (3\times 10^8)}{(1240\times 10^{-9})}$$
$$E_{\rm emitted}=\bf 8.2825\times 10^{-19}\;\rm J=\color{blue}{\bf 1.0}\rm \;eV$$
Now we know that the atom starts from the ground state at which energy is -4.0 eV and emits a photon of 1.0 eV. According to the known and the given graph, an electron must have been excited to an energy level of -1.0 eV from the ground level before the photon is emitted.
This means that the absorbed photon excited the atom from the ground state at -4.0 eV to -1.0 eV.
Hence, the energy of the absorbed photon is then given by
$$E_{\rm absorbed }=-1.0 - (-4.0)=\color{red}{\bf 3.0}\;\rm eV$$
$$\color{blue}{\bf [c]}$$
The atom here absorbers the same energy from the collision as it did in part b above since it was in the same ground state and emitted the same photon wavelength.
So the energy absorbed by the atom due to the collision is as same as the energy of the absorbed photon in part b above.
Thus, the energy transferred to the atom by the colliding electron is given by
$$E_{\rm collision}=\bf 3.0\;\rm eV$$
Therefore, the initial kinetic energy of the electron must have been at least 3.0 eV to excite the atom to the level of $n=3$.
If we assumed that our atom was at rest, then the initial kinetic energy of the electron is then given by
$$E_{\rm e}=\color{red}{\bf 3.0}\;\rm eV$$
