An oxide of iron contains 69.9% iron, its empirical formula, is:
(Given: Molar mass of Fe and O are 56 and 16 g mol$$^{-1}$$ respectively.)

The oxide contains (69.9%) iron by mass. Therefore, the percentage of oxygen is

$$100-69.9=30.1%.$$

Assuming a (100\ \text{g}) sample,

$$\text{Mass of Fe}=69.9\ \text{g},$$

$$\text{Mass of O}=30.1\ \text{g}.$$

The number of moles of each element is

$$\text{Moles of Fe}=\frac{69.9}{56}\approx1.248,$$

$$\text{Moles of O}=\frac{30.1}{16}\approx1.881.$$

Dividing both values by the smaller number of moles,

$$\frac{1.248}{1.248}=1,$$

$$\frac{1.881}{1.248}\approx1.5.$$

Thus, the simplest ratio is

$$Fe:O=1:1.5.$$

Multiplying both terms by (2) to obtain whole numbers,

$$Fe:O=2:3.$$

Hence, the empirical formula of the oxide is

$$Fe_2O_3.$$

Therefore, the correct answer is (Fe_2O_3).