Should I count them manually?

First we note the prime factorization $9955 = 5 \times 11 \times 181.$

So every positive integer that is not a positive multiple of any of $5,11$ or $181$ will be coprime to $9955.$ Assuming for the moment that the $80$ th such coprime integer is less than $181,$ we are looking for the least positive integer $N$ such that

$N - \left(\lfloor \dfrac{N}{5} \rfloor + \lfloor \dfrac{N}{11} \rfloor - \lfloor \dfrac{N}{55} \rfloor\right) = 80.$

As a means of expediting a solution, suppose for the moment that $N$ is divisible by $55.$ Then this equation becomes

$N - \left(\dfrac{N}{5} + \dfrac{N}{11} - \dfrac{N}{55}\right) = 80 \Longrightarrow \dfrac{40N}{55} = 80 \Longrightarrow N = 110.$

So there are $80$ integers less than or equal to $110$ that are coprime to $9955.$ But since $110$ itself is not coprime to $9955,$ we can conclude that $\boxed{109}$ is in fact the $80$ th smallest positive integer that is coprime to $9955.$