In my opinion , it is decidable to conclude whether L2 is reducible to L1 or not.U can see the link :
https://en.wikipedia.org/wiki/Reduction_(complexity)
Reduction of one problem to another is nothing but transforming one problem to another.
If L2 <m L1 it means that if an algorithm exists for solving L1 efficiently , then it can be used as a subroutine to find whether L2 can be solved efficiently.Now we are given L1 is decidable , meaning there exists an algorithm for language(or problem) L1.Now if using this algorithm we are able to solve L2 efficiently , this means L1 is reducible to L2 else not reducible.
So we are able to conclude the reducibilty of L2 into L1 . Hence it should be a decidable problem i.e. we are able to decide whether L2 is reducible to L1.
Secondly for any 2 languages s.t. L2 <m L1 so these 3 categories of languages(easy languages) work from right to left i.e. if right one is known we can conclude about left one :
a) If L1 is P , L2 is P.(But converse is not necessarily true)
b) If L1 is NP , L2 is NP.
c) If L1 is decidable , L2 is decidable.
Using the point c) , we can conclude that L2 is decidable as well.