Answer
$$v = 7.27\times 10^6m/s$$
$$V = 105.25V$$
Work Step by Step
According to De Broglie, wavelength of particle moving at a velocity v is given by
$$\lambda = \frac{h}{mv}$$
where m is the mass of the particle.
$$\therefore v = \frac{h}{m\lambda}$$
$$v = \frac{6.626\times 10^{-34}}{9.109\times 10^{-31}
\times 100\times 10^{-12} } = 7.27\times 10^6m/s$$
Since, the velocity is only $2.5$% of speed of light so here relativistic mass would be approximately equal to the rest mass.
Say that to acquire such velocity, the electron is passed through a potential V.
Energy provided by Potential V = K.E of the electron
$$eV = \frac{1}{2}m_ev^2$$
$$V = \frac{1}{2e}m_ev^2$$
$$V = \frac{1}{2\times 1.602\times 10^{-19}}(9.109\times 10^{-31})(7.27\times 10^6)^2$$
$$V = 105.25V$$
