Is
$$L = \{ \langle M \rangle \mid M = (\{Q_1, Q_2, . . . , Q_{100}\}, \{0, 1\}, \{0, 1, \_\}, δ, Q_1, Q_2, Q_3) \text{ is a decider}\}$$
decidable?
I know
$$HALT_{TM}= \{ \langle M \rangle \mid M \text{ is a decider}\}$$
is not decidable, but in the case we are given a specific Turing Machine that has 100 states, alphabet $\{0,1\}$, tape alphabet $\{0, 1, \_\}$ (where _ is space), a transition function $\delta$, a start state $Q_1$, an accept state $Q_2$ and a reject state $Q_3$.
Checking that M is in the correct format is easy but is it possible to build a Turing Machine that decides $L$? I assume I can use the fact that $HALT_{TM}$ is undecidable to show that $L$ also is, but I am not sure how to proceed.
And I don't see how I could reduce $L$ to $A_{TM}$ to prove by contradiction that it is not decidable?
How should I proceed?