The correct order of electron affinity is:

First, we recall the definition: the electron affinity (EA) of an element is the energy released when one mole of gaseous atoms gains one mole of electrons, forming one mole of gaseous anions. Mathematically we write $$\text{X(g)} + e^- \rightarrow \text{X}^-(g) + \text{Energy released}.$$ A larger release of energy (i.e., a more negative numerical value) means a higher electron affinity.

Now we describe the general periodic trends. Across a period from left to right, nuclear charge increases while the added electron enters the same principal shell, so the electron is more strongly attracted and electron affinity usually becomes more negative. Down a group the outer electron is added to a shell with a higher principal quantum number, so the nucleus-electron attraction lessens because of increased distance and shielding; therefore electron affinity normally becomes less negative down the group.

However, there is an important exception between the second-period element $$\mathrm F$$ and the third-period element $$\mathrm{Cl}$$. Although $$\mathrm F$$ is above $$\mathrm{Cl}$$ in Group 17, its atomic radius is so small that the incoming electron experiences strong inter-electronic repulsion in the compact $$2p$$ subshell. This repulsion partially offsets the high nuclear attraction, so the net energy released on adding an electron to $$\mathrm F$$ is slightly less than that released for $$\mathrm{Cl}$$. In numerical terms:

$$\text{EA}(\mathrm{Cl}) \approx -349\;\text{kJ mol}^{-1}, \qquad \text{EA}(\mathrm{F}) \approx -328\;\text{kJ mol}^{-1}.$$

Thus $$\mathrm{Cl}$$ actually has the higher electron affinity, even though periodic descent would normally suggest the opposite.

Next we compare oxygen with fluorine. Oxygen lies immediately to the left of fluorine in Period 2. Because nuclear charge is one unit smaller in oxygen and the added electron still enters the same $$2p$$ subshell, the attraction is weaker and the released energy is lower. Therefore,

$$\text{EA}(\mathrm F) > \text{EA}(\mathrm O).$$

Combining the two comparisons we have the following order from highest to lowest electron affinity:

$$\mathrm{Cl} > \mathrm F > \mathrm O.$$ Hence, the correct answer is Option D.