Unblind a RSA blinded signature I have blinded $m = 16$ with $r= 13$, $e = 7$ and $n = 209$.
This resulted in $m' = 464$, which led to a $s' = 8$.

Now, I want to unblind this $s'$ by using $s = s' \times r^{-1}$ but im stuck at filling in the $r^{-1}$?
So let's first recall how blinded RSA works:

Select primes $p=11$ and $q=19$, calculate the modulus $n=pq=209$, choose a public exponent $e=7$. Now calculate the signature exponent $d\equiv e^{-1} \pmod{(p-1)(q-1)}$ ($d=103, (p-1)(q-1)=180$).

Now let's go to the message-dependant processing:
A message $M$ is represented as the integer $m=16$ which is blinded by first choosing a random integer $r=13$ and then calculating $m'=m*r^e \bmod n$ ($m'=46$). Finally the signature operation is carried out using $s'\equiv(m')^d\equiv(m*r^e)^d\equiv m^d*r^{ed}\equiv s*r \pmod n$($s'=8$). Finally calculate $s\equiv s'*r^{-1} \pmod n$. For this you need $r^{-1} \bmod n$($r^{-1}=193$). This yields the signature $s=81$.