Among the species given below, the spin-only magnetic moment is highest for
(Given : Atomic number of Ti $=22$, Mn $=25$, Fe $=26$ and Co $=27$)

The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. We determine $n$ for each complex:

For $[\text{Ti(H}_2\text{O})_6]^{3+}$:
Ti is in $+3$ oxidation state. The electronic configuration of Ti$^{3+}$ is $[\text{Ar}] 3d^1$.
Number of unpaired electrons, $n = 1$.
$\mu = \sqrt{1(1+2)} = \sqrt{3}$ BM.

For $[\text{Mn(CN)}_6]^{3-}$:
Mn is in $+3$ oxidation state. The electronic configuration of Mn$^{3+}$ is $[\text{Ar}] 3d^4$.
Since CN$^-$ is a strong field ligand, pairing of electrons takes place. The configuration in the octahedral field is $t_{2g}^4 e_g^0$.
Number of unpaired electrons, $n = 2$.
$\mu = \sqrt{2(2+2)} = \sqrt{8}$ BM.

For $[\text{Fe(CN)}_6]^{3-}$:
Fe is in $+3$ oxidation state. The electronic configuration of Fe$^{3+}$ is $[\text{Ar}] 3d^5$.
CN$^-$ is a strong field ligand, causing pairing. The configuration is $t_{2g}^5 e_g^0$.
Number of unpaired electrons, $n = 1$.
$\mu = \sqrt{1(1+2)} = \sqrt{3}$ BM.

For $[\text{Co(NH}_3)_6]^{3+}$:
Co is in $+3$ oxidation state. The electronic configuration of Co$^{3+}$ is $[\text{Ar}] 3d^6$.
NH$_3$ acts as a strong field ligand for Co$^{3+}$, causing complete pairing. The configuration is $t_{2g}^6 e_g^0$.
Number of unpaired electrons, $n = 0$.
$\mu = 0$ BM.

Comparing the values, $[\text{Mn(CN)}_6]^{3-}$ has the highest number of unpaired electrons ($n=2$) and thus the highest spin-only magnetic moment.

Answer: $[\text{Mn(CN)}_6]^{3-}$