Let common ratio be $r$ and the first term be $a$. Then $ar^2\leq100 \implies r^2\leq10$
It's quite easy if we consider $r$ to be a natural number.
For $r=2$, $10\leq a\leq25$ and for $r=3$, $10\leq a\leq11$.
We get $18$ values. The problem arises when $r$ is not a natural number. $40,60,90$ definitely fulfils my condition, but I haven't been able to think of a foolproof method to find the number of GP's
Hint: The common ratio must be a rational number. Express it as a fraction $r=p/q$ in lowest terms, that is, such that $p$ and $q$ do not have any common factors.
Your geometric series of three is then $(a, ap/q, ap^2/q^2)$. They are all integers, so what do you know about $a$ and $q^2$? This should limit the possibilities of $q$ quite much. You still have to go through several cases, but this should help.