First, it should be intuitive that L is not decidable.
Decidable means that you can can tell in finite amount of time if a word (in this case, coding of turing machine) is in L.
In m opinion, this should be one of the first things that should come in mind while solving these kinds of questions.
Note that there are infinitely many string that have $101$ as prefix (for example $101, 1011, 10111$, or in general, $101^i$ for $ i \geq 1$).
There are several problems with your solution.
First, the input for a TM for $L$ is $\langle M\rangle$, and not $\langle M, w\rangle$.
Second, what happens if $M$ doesn't halt on $w$? Your TM for $L$ also doesn't halt, and hence can't decide $L$.
You can show that $ L \in RE\setminus R $.
Show that $L \in RE $ by describing a TM $M'$ that recognizes $L$ (meaning, on input $\langle M\rangle$, if $ \langle M\rangle \in L$ then $M'$ accepts $\langle M\rangle$, otherwise $M'$ rejects or doesn't halt.
An easy way to show that $L\notin R$ is a reduction $f$ from the halting problem $HP$ to $L$, such that if $\langle M,w\rangle \in HP$ , then $f(\langle M,w\rangle) \in L$ (Hint: $f(\langle M,w\rangle)$ we accept any input if and only if $M$ stops on $w$).