Step 1: Recall that dipole moment is a vector sum, not just a count of polar bonds.
A molecule can have polar bonds and still show zero or a small net dipole moment if those bond dipoles are arranged so they cancel out or partly oppose each other.
Step 2: Check the geometry of $BF_3$.
$BF_3$ is trigonal planar with three identical B-F bonds at $120^{\circ}$ to each other. By symmetry these three equal vectors add up to exactly zero:
\[ \mu(BF_3) = 0 \, \text{D} \]
Step 3: Compare the pyramidal shapes of $NH_3$ and $NF_3$.
Both molecules are pyramidal with a lone pair on nitrogen, but the lone pair's contribution to the resultant works differently in each. In $NH_3$, hydrogen is less electronegative than nitrogen, so the N-H bond dipoles point towards nitrogen, roughly the same direction as the lone pair, and the two effects add up. In $NF_3$, fluorine is more electronegative than nitrogen, so the N-F bond dipoles point away from nitrogen, roughly opposite the lone pair, and the two effects partly cancel.
Step 4: Use the known experimental values as a check.
Measured dipole moments are about $1.47$ D for $NH_3$ and only about $0.24$ D for $NF_3$, confirming that the lone-pair-opposing-bond-dipole effect in $NF_3$ nearly cancels its net moment while $NH_3$ keeps a large one.
Step 5: Put all three in order.
\[ \mu(BF_3) = 0 \lt \mu(NF_3) \approx 0.24 \lt \mu(NH_3) \approx 1.47 \]
Step 6: Why the other orders fail.
Any option that places $NH_3$ below $NF_3$, or $BF_3$ anywhere but first, ignores either the perfect symmetry of $BF_3$ or the cancelling effect of fluorine's electronegativity in $NF_3$.
Final Answer:
\[ \boxed{BF_3 \lt NF_3 \lt NH_3} \]