de Broglie wavelength of a moving particle, having mass m and velocity v is given by
$\lambda =\frac{\text{h}}{\text{p}}=\frac{\text{h}}{\mathrm{mv}}\text{or}\text{m}=\frac{\text{h}}{\lambda \text{v}}$
For an electron
${\lambda }_{\text{e}}=\frac{\text{h}}{{\text{m}}_{\text{e}}{\text{v}}_{\text{e}}}=\text{or}{\text{m}}_{\text{e}}=\frac{\text{h}}{{\lambda }_{\text{e}}{\text{v}}_{\text{e}}}$

for particle

${\lambda }_{}=\frac{\text{h}}{{\text{m}}_{}{\text{v}}_{}}=\text{or}\text{m}=\frac{\text{h}}{\lambda {\text{v}}_{}}$
$\text{Given :}\frac{\text{v}}{{\text{v}}_e}=3\text{,}\frac{\lambda }{{\lambda }_e}=1·813\times 10^{-4}$
$$\begin{gathered} \text{Mass of the particle,} \\ \frac{\text{m}}{{\text{m}}_{\text{e}}}=\left(\frac{{\text{v}}_{\text{e}}}{\text{v}}\right)\left(\frac{{\lambda }_{\text{e}}}{\lambda }\right) \\ \text{m}={\text{m}}_{\text{e}}\left(\frac{{\text{v}}_{\text{e}}}{\text{v}}\right)\left(\frac{{\lambda }_{\text{e}}}{\lambda }\right) \end{gathered}$$
Substituting the values, we get
$\text{m}=\left(9·1\times 10^{-31}\text{kg}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{1·813\times 10^{-4}}\right)$
m = 1.67 × 10^–27 kg