Solution

In $${ NH }_{ 3 }$$, the net dipole moment is non-zero because of Pyramidal shape.
In $${ H }_{ 2 }O$$, the lone pairs dipole moment is in one side and it has more dipole moment among $$4$$ options.
In $${ BCl }_{ 3 }$$, it is trigonal planar.

Net dipole moment between $${ Cl }_{ 2 }$$ and $${ Cl }_{ 3 }$$ is
$${ \mu }_{ { net }_{ (2\& 3) } }=\sqrt { { \mu }^{ 2 }+{ \mu }^{ 2 }+2\mu .\mu .cos120° } \\ \quad \quad \quad \ =\sqrt { 2{ \mu }^{ 2 }+2{ \mu }^{ 2 }(-1)\left( \dfrac { 1 }{ 2 } \right) } \\ \quad \quad \quad \ \Rightarrow \mu \\ { \mu }_{ { net }_{ (2\& 3) } } =\mu $$

The two cancels and net dipole moment is zero.
In $$SO_{ 2 }$$, many lone pairs are present which doesn't let the dipole moment to zero