The sum $100$ is not possible if we stop when the sum first becomes greater than $100$. Maybe you meant "greater than or equal to $100$." But we solve the problem as it stands.
Look at where we are just before we "go over." Maybe we are at $100$. Then $101$, $102$, $103$ are equally likely.
Maybe we are at $99$. Again, $101$, $102$, $103$ are equally likely. Maybe we are at $98$; same thing. Maybe we are at $97$; same thing.
Maybe we are at $96$. Now $103$ next is impossible, but $101$, $102$ are equally likely.
Maybe we are at $95$. Then only $101$ is possible among the three.
It is certainly possible that the sum just before we go over is $95$ or $96$. So $101$ is the most likely of $101$, $102$, $103$. And $102$ is next.
If the problem meant to say we stop when our sum is $\ge 100$, the same reasoning shows $100$ is the most likely "first over" number of our four choices.
Added Let $p_{100}$, $p_{99}$, $p_{98}$,, up to $p_{95}$ be the probabilities that we are respectively at $100$, $99$, and so on down to $95$ just before we go over. These $p_k$ are not equal, and would be fairly messy to compute. But we don't need to know them. The probability that we end up at $103$ is $\frac{1}{6}\left(p_{100}+p_{99}+p_{98}+p_{97}\right)$. The probability that we end up at $102$ is $\frac{1}{6}\left(p_{100}+p_{99}+p_{98}+p_{97}+p_{96}\right)$, clearly bigger. And the probability we end up at $101$ is even bigger.