An electron of mass $$m$$ and a photon have same energy $$E$$. The ratio of wavelength of electron to that of photon is: ($$c$$ being the velocity of light)

The de Broglie wavelength of an electron with energy $$E$$ is $$\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$, since the kinetic energy $$E = \frac{p^2}{2m}$$ gives momentum $$p = \sqrt{2mE}$$.

For a photon with energy $$E$$, we have $$E = \frac{hc}{\lambda_p}$$, so $$\lambda_p = \frac{hc}{E}$$.

The ratio of wavelengths is $$\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2mE}}{hc/E} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{E}{c\sqrt{2mE}} = \frac{\sqrt{E}}{c\sqrt{2m}} = \frac{1}{c}\sqrt{\frac{E}{2m}}$$.

This can be written as $$\frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}$$, so the correct answer is option 2.