Let be the problem of accepting Epsilion problem.
R = {<M>| M is a TM which accept Epsilion}
Now, Consider one Undecidable problem
ATM = {<M,w>|M is a TM and w is string}
Now we reduce ATM problem to R'(complement of the given problem).
Let F be the Reduction function for that.
F = "On input M,w "
1. Construct TM M
M1 = "On input <M,w>
a. for each x
b. if(x != w) "reject"
c. if(x==epsilon) "reject"
d. else RUN M on W. if M accept then accept , M reject then reject else loop.
[ M1's language contain either w or nothing ]
3. RUN R TM with input<M1>
4. If R' accept then "accept"
6. If R' reject then "reject"
So, ATM <m R'
==> ATM' <m R
As ATM' is not Turing reconizable language so, R is not Turing reconizable.
[we know ATM is undecidable language but Turing Reconizable language so ATM' is not R.E]
So, R is not R.E.