A photon and an electron, each of 20 eV energy, move in free space....

For a photon, the energy $E$ and linear momentum $p_{Ph}$ are related by:

$p_{Ph} = \dfrac{E}{c}$

For an electron, the kinetic energy $E$ and linear momentum $p_e$ are related by:

$p_e = \sqrt{2mE}$

The ratio of the linear momentum of the electron to that of the photon is:

$\dfrac{p_e}{p_{Ph}} = \dfrac{\sqrt{2mE}}{\dfrac{E}{c}} = c \sqrt{\dfrac{2m}{E}}$

Given:
$E = 20 \text{ eV} = 20 \times 1.6 \times 10^{-19} \text{ J} = 32 \times 10^{-19} \text{ J}$
$m = 9 \times 10^{-31} \text{ kg}$
$c = 3 \times 10^8 \text{ m/s}$

Substituting the values into the ratio:

$\dfrac{p_e}{p_{Ph}} = 3 \times 10^8 \times \sqrt{\dfrac{2 \times 9 \times 10^{-31}}{32 \times 10^{-19}}}$

$\dfrac{p_e}{p_{Ph}} = 3 \times 10^8 \times \sqrt{\dfrac{18 \times 10^{-31}}{32 \times 10^{-19}}}$

$\dfrac{p_e}{p_{Ph}} = 3 \times 10^8 \times \sqrt{\dfrac{9}{16} \times 10^{-12}}$

$\dfrac{p_e}{p_{Ph}} = 3 \times 10^8 \times \left(\dfrac{3}{4} \times 10^{-6}\right)$

$\dfrac{p_e}{p_{Ph}} = \dfrac{9}{4} \times 10^2 = 2.25 \times 100 = 225$