Electron is faster; Ratio of speeds is $13.54: 1$

Mass of the electron, $m_{ e }=9.11 \times 10^{-31} kg$

Mass of the proton, $m_{ p }=1.67 \times 10^{-27} kg$

Kinetic energy of the electron, $E_{ Ke }=10 keV =10^{4} eV$

$=10^{4} \times 1.60 \times 10^{-19}$

$=1.60 \times 10^{-15} J$

Kinetic energy of the proton, $E_{K p}=100 keV =10^{5} eV =1.60 \times 10^{-14} J$

For the velocity of an electron $v_{e},$ its kinetic energy is given by the relation:

$E_{ Ke }=\frac{1}{2} m v_{ c }^{2}$

$\therefore v_{e}=\sqrt{\frac{2 \times E_{ Ke }}{m}}$

$=\sqrt{\frac{2 \times 1.60 \times 10^{-15}}{9.11 \times 10^{-31}}}=5.93 \times 10^{7} m / s$

For the velocity of a proton $v_{ p },$ its kinetic energy is given by the relation:

$E_{ Kp }=\frac{1}{2} m v_{ p }^{2}$

$v_{p}=\sqrt{\frac{2 \times E_{ Kp }}{m}}$

$\therefore v_{ p }=\sqrt{\frac{2 \times 1.6 \times 10^{-14}}{1.67 \times 10^{-27}}}=4.38 \times 10^{6} m / s$

Hence, the electron is moving faster than the proton.

The ratio of their speeds:

$\frac{v_{c}}{v_{p}}=\frac{5.93 \times 10^{7}}{4.38 \times 10^{6}}=13.54: 1$