Answer
See explanation.
Work Step by Step
We know the momentum of the free electron, so we can calculate the wavenumber k and the frequency .
$k=\frac{p}{\hbar}=-4.27\times10^{10}m^{-1}$
Now use that expression for k.
$\omega=\frac{\hbar k^2}{2m}=1.05\times10^{17}s^{-1}$
Now express its wave function.
$\Psi(x,t)=Ae^{ikx}e^{-i \omega t}$
$\Psi(x,t)=Ae^{-i(4.27\times10^{10}m^{-1})x}e^{-i (1.05\times10^{17}s^{-1}) t}$
