We can use Rice's Theorem to answer such question. Let's build up to it.
Let's say $P$ is a function, such that if a language $L$ satisfies some given property; then $P(L)=1$.
Otherwise, $P(L)=0$
This is a pretty intuitive Boolean notation.
Now, Rice's Theorem states that every non-trivial property of a Recursively Enumerable languages is undecidable.
Introduction to Automata Theory by Hopcroft and Ullman, 3rd Edition, page 397.
What is a non-trivial property of RE languages?
A property of RE languages is said to be non-trivial, if there exists Turing Machines $M_1$ and $M_2$, such that $P(L(M_1))=1$ and $P(L(M_2))=0$
In plain English, this means that the property of an RE language is non-trivial, if there exists some Turing Machine whose languages have that property, and there also exists some Turing Machine whose languages don't have that property.
Now let's use Rice's Theorem.
$L_1$ is undecidable because we can construct a TM that acceps nothing, and we can construct a TM that acceps something.
Similarly; $L_3$ and $L_4$ are also undecidable.
$L_2$ is decidable ONLY because of the fact that we've been mentioned the number of steps.
Otherwise it would be undecidable, too.
Each "step" in a Turing Machine would take a finite amount of time. Let's say it takes maximum $x$ units of time. So, for $L_2$ we only have to observe the Turing Machine for $100x$ units of time. We can't fall in infinite loop here; hence, decidable.
Option A