Use de-Broglie wavelength relation

The de-Broglie wavelength is $\lambda = \frac{h}{mv}$

Write wavelengths for the particle and the electron

For the particle: $\lambda_p = \frac{h}{m_p v_p}$

For the electron: $\lambda_e = \frac{h}{m_e v_e}$

Form the ratio

Given that the ratio of wavelength of particle to electron is $2:1$, $\frac{\lambda_p}{\lambda_e} = 2$

So, $\frac{h/(m_p v_p)}{h/(m_e v_e)} = 2$ $\frac{m_e v_e}{m_p v_p} = 2$

Use the speed condition

The particle travels $4$ times as fast as the electron: $v_p = 4v_e$

Substitute into the ratio: $\frac{m_e v_e}{m_p (4v_e)} = 2$ $\frac{m_e}{4m_p} = 2$

Solve for the particle mass

$m_e = 8m_p$ $m_p = \frac{m_e}{8}$

Match with the options

The mass of the particle is $\boxed{\frac{1}{8} \text{ times the mass of electron}}$

So the correct option is D.