$L_2 - L_1 = L_2 \cap {L_1}^c$
Now, complement of a recursive language is always recursive - so ${L_1}^c$ is recursive and hence recursively enumerable too.
Intersection of two recursively enumerable languages always gives a recursively enumerable language (Recursively Enumerable set is closed under union and intersection but not under complement).
So, $L_2 - L_1$ is guaranteed to be recursively enumerable.