Answer
$3$ $MeV$
Work Step by Step
Total energy of the deuteron= kinetic energy of the deuteron+potential energy of the deuteron
$\implies E=K+U$
If the deuteron reflects the barrier, the value of $U=0 $ on the reflected side of the barrier because the the height of potential barrier is zero in this region.
Therefore, $K=E=3$ $MeV$
$\therefore$ If the deuteron reflects the barrier, the kinetic energy of the deuteron remains $3$ $MeV$
