The spin-only magnetic moment ($\mu$) is given by the formula $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. For a magnetic moment of 1.73 BM, the number of unpaired electrons must be $n=1$ (since $\sqrt{1(1+2)} = \sqrt{3} \approx 1.73$).

1. $[Cu(NH_3)_4]^{2+}$: The central metal ion is Cu in the +2 oxidation state. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. It has 1 unpaired electron ($n=1$). Thus, its spin-only magnetic moment $\mu = 1.73$ BM .
2. $[Ni(CN)_4]^{2-}$: The central metal ion is Ni in the +2 oxidation state ($[Ar] 3d^8$). The cyanide ion ($CN^-$) is a strong field ligand, which causes the pairing of the two unpaired $d$ electrons. It has 0 unpaired electrons ($n=0$), making it diamagnetic. Thus, $\mu = 0$ BM .
3. $TiCl_4$: The central metal ion is Ti in the +4 oxidation state ($[Ar] 3d^0$). It has no $d$ electrons and hence 0 unpaired electrons ($n=0$). Thus, $\mu = 0$ BM.
4. $[CoCl_6]^{4-}$: The central metal ion is Co in the +2 oxidation state ($[Ar] 3d^7$). The chloride ion ($Cl^-$) is a weak field ligand, so no pairing of electrons takes place. It has 3 unpaired electrons ($n=3$). Thus, $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM .

Therefore, $[Cu(NH_3)_4]^{2+}$ is the complex that shows a magnetic moment of 1.73 BM.