Answer
$1.03\times10^6m/s$
Work Step by Step
The kinetic energy of the ejected electrons becomes potential energy at the top plate. Energy is conserved.
$$KE_i=PE_f$$
$$\frac{1}{2}mv^2=qV$$
$$v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(-1.60\times10^{-19}C)(-3.02V)}{9.11\times10^{-31}kg}}$$
$$=1.03\times10^6m/s$$