Answer
$2\;eV$
Work Step by Step
Total energy of the electron in the left side region ($V_1$ region) is given by
$E_1=\text{Kinetic energy}+\text{Potential energy}$
or, $E_1=2+9\;eV=11\;eV$
If the electron is trapped in $V_2$ region, it should have to loose its energy by emitting a photon.
According to given energy diagram, to barely become trapped, the energy of the electron must has to be $E_2=9\;eV$
Therefore, the energy of the emitted photon is
$E=E_2-E_1$
or, $E=(11-9)\;eV=2\;eV$
