Answer
${\bf 6.2\times 10^5}\;\rm m/s$
Work Step by Step
The electron initially before the collision was moving at a speed of $1.4\times 10^6$ m/s.
After the collision by some time, the atom emits a photon with a wavelength of 1240 nm, which we can find its energy by
$$E_{\rm photon}=\Delta E_{\rm atom}=\dfrac{hc}{\lambda}$$
Plug the known;
$$ \Delta E_{\rm atom}=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(1.6\times 10^{-19})(1240\times 10^{-9})}=\bf 1.0\;\rm eV$$
This means that the atom loses 1 eV in this process which is the same energy of the emitted photon. This transition could be found from $1\rightarrow 2$ or $1\rightarrow 3$ since we found the wavelengths of these transitions in the previous problem and they were 276 nm and 355.
So the only remaining transition is $3\rightarrow 2$.
This means that the collision gives the atom energy that makes it move from the ground state to $n=3$ and then it loses one photon and moves to $n=2$.
So, the energy needed to move this atom from ground state to $n=3$ is given by
$$ \Delta E_{\rm atom}=E_3-E_1=-2-(-6.5)=\bf 4.5\;\rm eV$$
This 4.5 eV is the exact amount of energy the electron loses in this collision.
$$\Delta K_{\rm electron}=K_i-K_f=4.5 \;\rm eV$$
where $K=\frac{1}{2}mv^2$, so
$$ \frac{1}{2}m_ev_i^2-\frac{1}{2}m_ev_f^2=4.5 \;\rm eV$$
Solving for $v_f$, and converting eV to J.
$$ v_i^2- v_f^2=\dfrac{2(4.5) (1.6\times 10^{-19})}{m_e }$$
$$v_f=\sqrt{v_i^2-\dfrac{2(4.5) (1.6\times 10^{-19})}{m_e }}$$
Plug the known;
$$v_f=\sqrt{(1.4\times 10^6)^2-\dfrac{2(4.5) (1.6\times 10^{-19})}{(9.11\times 10^{-31})}}$$
$$v_f=\color{red}{\bf 6.2\times 10^5}\;\rm m/s$$
