The correct order of dipole moment of the different methyl halides is:

\(CH_3I > CH_3Br > CH_3Cl > CH_3F\)

The correct answer is option 3. \(CH_3I > CH_3Br > CH_3Cl > CH_3F\).

The dipole moment of a molecule is a measure of its polarity, which is determined by the difference in electronegativity between the atoms and the molecular geometry.

In methyl halides (\(CH_3X\), where \(X\) is a halogen), the dipole moment depends on the electronegativity of the halogen atom (\(X\)). Fluorine (F) is the most electronegative halogen, followed by chlorine (Cl), bromine (Br), and iodine (I).

Let us analyze the provided correct order of dipole moments:

\[ CH_3I > CH_3Br > CH_3Cl > CH_3F \]

1. \(CH_3I\) (methyl iodide): Iodine (I) is the least electronegative halogen among fluorine, chlorine, bromine, and iodine. In \(CH_3I\), the carbon-iodine (\(C-I\)) bond has the largest difference in electronegativity, resulting in the largest dipole moment among the methyl halides.

2. \(CH_3Br\) (methyl bromide): Bromine (Br) is less electronegative than iodine (I) but more electronegative than chlorine (Cl) and fluorine (F). Therefore, \(CH_3Br\) has a larger dipole moment than \(CH_3Cl\) and \(CH_3F\), but smaller than \(CH_3I\).

3. \(CH_3Cl\) (methyl chloride): Chlorine (Cl) is more electronegative than bromine (Br) and iodine (I) but less electronegative than fluorine (F). Thus, \(CH_3Cl\) has a smaller dipole moment than \(CH_3Br\) and \(CH_3I\), but larger than \(CH_3F\).

4. \(CH_3F\) (methyl fluoride): Fluorine (F) is the most electronegative halogen. Therefore, \(CH_3F\) has the smallest dipole moment among the methyl halides.

Therefore, the correct order of dipole moments is as provided:

\[ CH_3I > CH_3Br > CH_3Cl > CH_3F \]

This order reflects the trend in electronegativity of the halogens, where the larger the difference in electronegativity between carbon and halogen, the larger the dipole moment.