Assume that the probability of improper documentation is indeed $0.2$. Let $X$ be the number of improperly documented vouchers in a random sample of $100$. On the assumption that the number of vouchers we are sampling from is very large, we can assume that $X$ has binomial distribution, with "$n$" equal to $100$, and "$p$" equal to $0.2$.
Then the probability that exactly $k$ vouchers are improperly documented is $\dbinom{100}{k}(0.2)^k(0.8)^{100-k}$.
The probability that $31$ or more vouchers are improperly documented is $\sum_{k=31}^{100}\binom{100}{k}(0.2)^k(0.8)^{100-k}.$ Nowadays, one can calculate this without much trouble, with software free or expensive.
In the old days, direct calculation was a lengthy process. So the fact that $X$ has distribution well approximated by the normal would be used. The mean of $X$ is $(100)(0.2)=20$. The variance of $X$ is $(100)(0.2)(0.8)=16$, so $X$ has standard deviation $4$. The standard deviation is very simple, indication perhaps that you are expected to use a normal approximation.
The probability that a normal with mean $20$ and standard deviation $4$ is $\ge 31$ is easy to calculate using tables of the standard normal.
I do not know whether you are expected to use the continuity correction. If we don't, then the probability that $X\le 30$ is approximately the probability that $Z\le \frac{30-20}{4},\tag{$1$}$ where $Z$ is standard normal. Then our estimate is $1$ minus the number obtained in $(1)$.
If we use the continuity correction, the $30$ in $(1)$ is replaced by $30.5$.