The most polar compound among the following is:

Polarity of a molecule is expressed by its net dipole moment $$\mu$$ (Debye, D). The vector $$\mu$$ is obtained by adding the individual bond-dipoles of the molecule. Two factors govern $$\mu$$:

(i) Magnitude of each bond dipole  $$\propto$$  electronegativity difference $$\Delta\chi$$ between the bonded atoms.
(ii) Spatial arrangement of the bonds (molecular geometry); symmetric arrangements can make the bond vectors cancel.

Let us analyse the given compounds one by one.

HF is a diatomic molecule, so the bond dipole is the net dipole. Electronegativity difference $$\Delta\chi = \chi_{F} - \chi_{H} \approx 3.98 - 2.20 = 1.78$$ produces a dipole moment $$\mu = 1.82\;{\rm D}$$.

HCl is also diatomic. The greater bond length (136 pm vs 92 pm in HF) weakens the dipole in spite of a sizable $$\Delta\chi$$, so $$\mu = 1.08\;{\rm D}$$.

H2O is bent (bond angle $$104.5^{\circ}$$). Each O-H bond dipole (about $$1.52\;{\rm D}$$) adds vectorially. Because they are not collinear, cancellation is incomplete, and the resultant is large:
$$\mu_{\text{H}_2\text{O}} = 2 \times (1.52\;{\rm D})\cos\!\left(\frac{104.5^{\circ}}{2}\right) \approx 1.85\;{\rm D}$$.

CO2 is linear (O-C-O angle $$180^{\circ}$$). The two identical C=O bond dipoles are equal and opposite; they cancel completely, so $$\mu = 0\;{\rm D}$$ (non-polar).

Comparing the calculated dipole moments:
$$\mu(\text{HF}) = 1.82\;{\rm D}$$
$$\mu(\text{HCl}) = 1.08\;{\rm D}$$
$$\mu(\text{H}_2\text{O}) = 1.85\;{\rm D}$$
$$\mu(\text{CO}_2) = 0\;{\rm D}$$

The largest value is for water. Therefore the most polar compound in the list is H2O.

Option C which is: H2O.