The de-Broglie wavelength of a photon is twice, the de-Broglie wavelength of an electron. The speed of the electron is $v_e=\frac{c}{100}$Then,

Suppose, Mass of electron$=m_e,$  Mass of photon$=m_p,$  
Velocity of electron$=v_e$  and velocity of photon$=v_p$  Thus, for electron, de-Broglie wavelength

             $$\begin{gathered} {\lambda }_e=\frac{h}{m_e{v}_e} \\ =\frac{h}{m_e(c/100)}=\frac{100h}{m_{e}c}\text{ (given) (i)} \end{gathered}$$

Kinetic energy,   $E_e=\frac{1}{2}{m}_e{v}_e^2$

$$\begin{gathered} \Rightarrow m_e{v}_e=\sqrt{2E_e{m}_e} \\ so, {\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{\sqrt{2m_e{E}_e}} \\ E_e=\frac{h^2}{2{\lambda }_e^2{m}_e} \left(ii\right) \end{gathered}$$

For photon of wavelength ${\lambda }_p,$ energy

$$\begin{gathered} E_p=\frac{hc}{{\lambda }_p}=\frac{hc}{2{\lambda }_e} \left[\because {\lambda }_p=2{\lambda }_e\right.] \\ \therefore \frac{E_p}{E_e}=\frac{hc}{2{\lambda }_e}\times \frac{2{\lambda }_e^2{m}_e}{h^2} \\ =\frac{{\lambda }_e{m}_{e}c}{h}=\frac{100h}{m_{e}c}\times \frac{m_{e}c}{h}=100 \end{gathered}$$

$$\begin{gathered} so, \frac{E_e}{E_p}=\frac{1}{100}={10}^{-2} \\ For electron, p_e=m_e{v}_e=m_e\times c/100 \\ \frac{p_e}{m_{e}c}=\frac{1}{100}={10}^{-2} \end{gathered}$$