M is a Determinstic TM with one tape, c2 is c1 reachable, if it's reachable within finite positive time
Since no answer has been posted I will attempt the answer my self:
Since Giacomo explained why $L$ is TR(I thank him for that) I won't talk about it.

I will prove $L$ is not decidable:

We do that by building a machine $K$ that will decide the halting problem $L_{1}$ using $G$ the decider of $L$ that we assume exists.

$L_{1}$={&lt; $R$, $w$ > | $R$ is a TM; $w \in \Sigma^{*} $; $R$ halts on $w$ }

Description of $K$:

Build TR $H$ that has a configuration $c$ as follows:


"Configuration $c$": On input &lt;$R$, $w$ > if $R$ halts on $w$ go to configuration $d$ which on any input goes back to configuration $c$.
else go to configuration $e$ and loop forever.


Run $G$ on input &lt;$H$,$c$ > if $G$ accepts, accept else reject.(if $H$ accepts $K$ accepts and vice-versa)

Giving &lt; $R$,$w$> to $K$ as input, $K$ will decide if $R$ halts on $w$  if and only if (iff) $G$ decides if $c$ is reachable from $c$ in $H$.

But we know that the halting problem is not decidable in contradiction to what we have assumed in the first place, so $G$ cannot exist there for $L$ is not decidable.