Dipole moment of (1s,2s,3s)-1,2,3-trichlorocyclopropane
The issue stems from the fact that you may not have understood what each bond notation means.
Here, the solid wedge indicates that the bond is toward us and so a 3-D view of the same molecule would look as follows:

So as you can see the dipole moments – instead of cancelling, add up. So simply by elimination option (d) is the right answer
For trans-1,3-dichlorocyclobutane, the reason for the zero dipole moment is explained in ron's answer to the question &quot;Does trans 1,3-dichlorocyclobutane have zero dipole moment?&quot;.


[...] Conformers A and B both have $C_$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers A and B are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.

Therefore (d) is the correct answer and not (c).