Answer should be A.
$L_{ne}$ says the input encoded TM $M$ should accept a string - language of $M$ is not empty set. i.e., if we get a string which is accepted by $M$ then our work is done - we get an answer for the yes part. But in case we do not get a string we have to check all infinite strings to know finally whether it will be not equal to $\phi$ . Therefore we do not have answer for no part. Hence it is RE but not recursive.
$L_e$ is not even re because we do not have answer for yes part . We have to check all the infinite strings to answer whether it finally accepts $\phi$ .
We can also apply Rice's theorem here. Both $L_{ne}$ and $L_{e}$ describes non-trivial properties of r.e. languages. So, as per Rice's theorem both are undecidable. Now, $L_{e}$ is a non-monotonic property ($L(T_{yes}) \subset L(T_{no})$ possible for empty set and any non-empty set respectively). So, $L_{e}$ is not even r.e. as per Rice's theorem part 2. Rice's theorem part 2 cannot be used for $L_{ne}$ since it is not a non-monotonic property (we cannot get any $T_{yes}, T_{no}$ such that $L(T_{yes} \subset L(T_{no})$. So, we cannot say if it is not r.e. using Rice's theorem and we have to use the first method for this part.
http://www.gatecse.in/rices-theorem/