Here, work function $\omega_0 = 3\,eV$ and photon energy $E_0 = 2\,eV$
As, Einstein’s photoelectric equation $E_0 = W_0 + KE_0$
So, $KE_0 = E_0 - W_0$
$\Rightarrow KE_0 = 2 - 3 = -1 \,eV$
Since, kinetic energy is a scalar quantity i.e., $KE > 0$, positive.
So, the photoelectron does not ejected and $KE = 0$