Paramagnetic behaviour is directly proportional to the number of unpaired electrons in the central metal ion. Let us calculate the number of unpaired electrons ($n$) for each complex:

1. In $[Zn(NH_3)_6]^{2+}$, $Zn$ is in the $+2$ oxidation state. Its electronic configuration is $[Ar] 3d^{10}$. All electrons are paired, so $n = 0$.
2. In $[Ti(NH_3)_6]^{3+}$, $Ti$ is in the $+3$ oxidation state. Its electronic configuration is $[Ar] 3d^1$. Thus, $n = 1$.
3. In $[Cr(NH_3)_6]^{3+}$, $Cr$ is in the $+3$ oxidation state. Its electronic configuration is $[Ar] 3d^3$. These three electrons occupy the lower energy $t_{2g}$ orbitals singly ($t_{2g}^3 e_g^0$). Thus, $n = 3$.
4. In $[Co(NH_3)_6]^{3+}$, $Co$ is in the $+3$ oxidation state. Its electronic configuration is $[Ar] 3d^6$. Ammonia ($NH_3$) acts as a strong field ligand with $Co^{3+}$, causing the pairing of all six $d$-electrons in the $t_{2g}$ level ($t_{2g}^6 e_g^0$). Thus, $n = 0$ (diamagnetic). Since $[Cr(NH_3)_6]^{3+}$ has the maximum number of unpaired electrons ($n=3$), it exhibits the highest paramagnetic behaviour.