Let us examine the dipole moment of each molecule:

• $\text{CO}_2$: It has a linear structure. The bond dipoles of the two $\text{C=O}$ bonds are equal and opposite, canceling each other out, resulting in a net dipole moment of zero.
• $\text{CH}_4$: It has a highly symmetrical tetrahedral geometry. The vector sum of the four $\text{C-H}$ bond dipoles is zero, so it is a non-polar molecule.
• $\text{NH}_3$ and $\text{NF}_3$: Both have a trigonal pyramidal shape with one lone pair on the central nitrogen atom. In $\text{NH}_3$, the orbital dipole due to the lone pair is in the same direction as the resultant dipole moment of the three $\text{N-H}$ bonds, leading to a higher net dipole moment ($4.90 \times 10^{-30} \text{ C m}$ or $1.47 \text{ D}$). In $\text{NF}_3$, the orbital dipole is in the direction opposite to the resultant dipole moment of the three $\text{N-F}$ bonds, which decreases the net dipole moment ($0.8 \times 10^{-30} \text{ C m}$ or $0.23 \text{ D}$). Therefore, $\text{NH}_3$ has the maximum dipole moment among the given options .