Answer
a) ${\bf 2.06\times 10^6}\;\rm m/s$
b) ${\bf12.09} \;\rm V$
Work Step by Step
$$\color{blue}{\bf [a]}$$
To induce the emission of 656 nm light from the $3\rightarrow2$ transition, the atom must first be excited from the ground state to the $n=3$ level.
Let's assume that the entire kinetic energy of the electron before the collision is transferred to the hydrogen atom.
Now we can determine the minimum speed of the electron beam by calculating the minimum energy needed to excite the atom from the ground state to $n=3$.
$$\Delta E_{1\rightarrow 3}=E_3-E_1=-1.51-(-13.6)=\bf12.09\;\rm eV $$
This is the energy required to excite the atom to the $n=3$ state which is the minimum energy needed from the electron.
Thus, the minimum kinetic energy of the electron before the collision is given by
$$K_{\rm min}=\frac{1}{2}mv_{\rm min}^2=12.09\;\rm eV$$
Hence,
$$v_{\rm min}=\sqrt{\dfrac{2(12.09)(e)}{m}}$$
Plug the known;
$$v_{\rm min}=\sqrt{\dfrac{2(12.09)(1.6\times 10^{-19})}{9.11\times 10^{-31}}}$$
$$v_{\rm min}=\color{red}{\bf 2.06\times 10^6}\;\rm m/s$$
$$\color{blue}{\bf [b]}$$
Recalling that the kinetic energy of the electron is given by
$$K=e(\Delta V)$$
So,
$$\Delta V=\dfrac{K}{e}=\dfrac{12.09\;e}{e}=\color{red}{\bf12.09} \;\rm V$$
