How this language belong to R?
The language $L$ is not decidable, this is an immediate consequence of Rice's theorem (you can instead try and prove that $L$ is not decidable directly using reductions). To begin with, there are two machines $M_1$ and $M_2$ such that $L(M_1)\in coRE$ and $L(M_2)\notin coRE$. For example, $M_1$ can be a machine that accepts all inputs, and $M_2$ can be a machine that recognizes the language $Halt_{TM} = \{\langle M, w\rangle: \text{ $M$ halts on $w$ }  \}$ which is known to be in $RE\setminus coRE$. So the property $P = \{ M: \text{$M$ is a TM, and $L(M)\in coRE$}\}$ is a non-trivial property. Also, for every two machines $M_1$ and $M_2$ with $L(M_1) = L(M_2)$, it holds that $L(M_1)\in coRE$ iff $L(M_2)\in coRE$. Thus, either both $M_1$ and $M_2$ are in $P$, or both are not in $P$. Hence, the property $P$ is also semantic, and so by Rice's theorem $L$ is not decidable.
Maybe you saw the claim suggesting that $L' = \{ \langle M\rangle:  \text{ $M$ is a TM, and $L(M) \in RE$}  \}$ is decidable, and this follows simply from the fact that $L'$ is the language of all legal encodings of TMs.