Answer is (3)
Initial de-broglie wavelength will be,
$\lambda_0=\frac{h}{mV_0}$ …(i)
Acceleration of electron will be,
$a=\frac{eE_0}{m}$
Velocity after time t,
$\mathbf{V}=\left(\mathbf{V}_{0}+\frac{\mathrm{eE}_{0}}{\mathrm{~m}} \mathrm{t}\right)$
So, $\lambda=\frac{h}{m V}=\frac{h}{m\left(v_{0}+\frac{e E_{0}}{m} t\right)}$
$=\frac{h}{m V_{0}\left[1+\frac{e E_{0}}{m V_{0}} t\right]}=\frac{\lambda_{0}}{\left[1+\frac{e E_{0}}{m V_{0}} t\right]}$…(ii)
Divide (ii) by (i) we get,
$\lambda=\frac{\lambda_{0}}{\left[1+\frac{\mathrm{e} \mathrm{E}_{0}}{\mathrm{~m} \mathrm{~V}_{0}} \mathrm{t}\right.}$