Answer
${\bf 658}\;\rm nm$
Work Step by Step
The electron initially before the collision was moving at a speed of $2.1\times 10^6$ m/s.
This means that the initial kinetic energy of the electron is
$$K=\frac{1}{2}mv^2=\frac{1}{2}(9.11\times10^{-31})( 2.1\times 10^6)^2=\bf 2.0\times 10^{-18}\;\rm J$$
$$K= \bf 12.55\;\rm eV$$
If the orbital electrons absorb all the energy of the bullet electron after the collision, it can go from the ground level up to $n=3$ at which its energy difference is 12.09 eV.
This means that it can't go to $n=4$ at which its energy difference is 12.75 eV.
Now we know that after a while the atom then undergoes a quantum jump with $\Delta n=1$, which means that it moves down from $n=3$ to $n=2$.
The energy of this transition is given by
$$\Delta E=E_2-E_3=\dfrac{hc}{\lambda}$$
where $\lambda$ is the wavelength of the emitted photon due to this transition
Hence,
$$\lambda=\dfrac{(6.63\times 10^{-34})(3\times 10^8)}{(-1.51-[-3.4])(1.6\times 10^{-19})}=\color{red}{\bf 658}\;\rm nm$$
