If you're using the normal approximation, you're going to use n, p, and q from your binomial distribution to calculate a mean and standard deviation that you can use with a z-score. That is what 'normal approximation' means.
So, in your case:
n = 183
p = 0.22 (because 22% voted)
q = 1-0.22 = 0.78 (the % who didn't vote)
mean = n*p = 183*0.22 = 40.26
standard deviation = sqrt(n*p*q) = sqrt(183*0.22*0.78) = 31.4028
Now, you're ready to plug that into a z-score formula. :) Do you know what to do next?