As an opening comment, which you should not use to try to influence your management (especially given that you are an intern), Type I service level is almost always a terrible objective. What it measures is the probability that you suffer no stockouts over the leadtime, but, from a business perspective, what you'd like to measure is, far more often, the expected number of stockouts over the order cycle $/$ the expected demand over the order cycle, i.e., the fraction of demands that go unfilled. In some situations, e.g. stocking a helicopter for rescue missions where an out of stock could mean permanent disability or death, Type I service level is appropriate, but not in the usual business case. I won't go into more detail, since it's off-topic, but include it here for your future information.
The Type I service level given an order point of $s$ is nothing more than the probability of seeing fewer than $s$ demands over leadtime. As such, if you have a probability distribution of leadtime demand $F(x)$, and a target service level $\alpha$, the corresponding order point $s_{\alpha}$ is:
$$s_{\alpha} = F^{-1}_{LTD}(\alpha)$$
where $LTD$ refers to "leadtime demand". In the case of the Normal distribution, it so happens that this is the same as:
$$s_{\alpha} = \mu_{LTD} + \sigma_{LTD}*z_{\alpha}$$
where $\mu$ is the mean demand, and $\sigma$ is the standard deviation of demand. Where does $z_{\alpha}$ come from? It's the inverse of the standard Normal distribution's cumulative density function at $\alpha$:
$$z_{\alpha} = F^{-1}(\alpha | \mu=0, \sigma=1)$$
and
$$F^{-1}(\alpha | \mu_{LTD}, \sigma_{LTD}) = \mu_{LTD} + \sigma_{LTD} F^{-1}(\alpha|\mu=0,\sigma=1) $$
So using $z_{\alpha}$ with the Normal distribution is the same as calculating the inverse of the cumulative distribution of leadtime demand, where leadtime demand is Normally distributed.
When you aren't working with the Normal distribution, you won't typically have any equivalent formula to make life appear simpler. This is true of both the Poisson and Negative Binomial distributions. In the case of these two distributions, it's most straightforward just to calculate order point from the initial equation, using the appropriate parameters for the leadtime demand distribution in question.
ETA:
For example, assume your daily demand has mean $0.2$ units and standard deviation $0.6$ units, and you have a leadtime of one week. Then $\mu_{LTD} = 1.4$ and $\sigma_{LTD} = 1.59$. The parameters of the Negative Binomial distribution are $r = 1.75$ and $p = 0.556$.
If you want to stock to a Type I service level of $95\%$, you'd solve for:
$$s_{\alpha} = F^{-1}_{LTD}(\alpha) = F^{-1}(\alpha=0.95;r=1.75,p=0.556) = 5$$
and your order point would be $5$.
Best Answer
Exact Poisson: First adjust the rate to 240 to match 12 hrs. Then let $X$ be the number of calls in 12 hrs, where $X \sim \mathsf{Pois}(240)$ and you seek $P(X = 250) = 0.0205.$ (Computation in R, but computation using the Poisson PDF, or PMF, isn't difficult on a calculator.)
Normal approximation: You have $\mu = E(X) = 240$ and $\sigma^2 = Var(X) = 240.$ So $\sigma = \sqrt{240} = 15.4919.$ So $$P(X = 240) \approx P(249.5 < Y < 250.5),$$ where $Y \sim \mathsf{Norm}(\mu = 240, \sigma=15.4919).$ Below is the computation of this normal approximation using R. I suppose you are expected to get the answer by standardizing and using normal tables. See how close you can get. [It may not be exact because a little rounding may be necessary to use printed normal tables.]
Binomial approximation. Don't give up on using a binomial distribution. When $n$ is very large and $p$ is small, $\mathsf{Binom}(n, p)$ is well approximated by $\mathsf{Pois}(\lambda = np).$ Let's try $n = 2400$ and $p = 1/10.$
Or, maybe better, $n = 24,000$ and $p = 1/100.$
Binomial probabilities can be a little messy to compute on a calculator because the factorials in the binomial coefficient are so large.
The plot below shows the Poisson distribution (black bars, values between 230 and 260), the approximating normal density curve (blue), and the second binomial approximation (purple circles).
Note: In any case, it is useful to know relationships among binomial, Poisson, and normal distributions.