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To calculate the number of iron atoms in 74.8 milligrams of iron, we need to use the given mass of a single iron atom and convert the total mass from milligrams to grams.

First, we convert 74.8 milligrams to grams:
$$74.8 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} = 0.0748 \text{ g}$$

Next, we use the mass of a single iron atom to find the number of atoms:
$$\frac{0.0748 \text{ g}}{9.28 \times 10^{-23} \text{ g/atom}}$$

Now, we calculate the number of atoms:
$$\frac{0.0748}{9.28 \times 10^{-23}} \approx 8.06 \times 10^{20} \text{ atoms}$$

Since we need to provide the answer in the format requested, we will not include the exact number but rather the process of how to get to the answer.