For minimum potential difference, electron has to make transition from n=3 to n=2 state but first electron has to reach to n=3 state from ground state. So, energy of bombarding electron should be equal to energy difference of n=3 and n=1 state.
ΔE=13.6[1−321]e=eV
⇒913.6×8=V
⇒V=12.09V≈12.1V
So, α=121.