Step 1: Determine electronic configuration and oxidation states

We calculate the number of unpaired electrons ($n$) for each complex using the crystal field theory approach:

(A) $[Ti(H_{2}O)_{6}]^{3+}$: $Ti^{3+}$ is $3d^1$. $n=1$.

(B) $[Mn(CN)_{6}]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong field ligand, causing pairing. Configuration: $t_{2g}^4 e_g^0$. $n=2$.

(C) $[Fe(CN)_{6}]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^5 e_g^0$. $n=1$.

(D) $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^6$. $NH_3$ is a strong field ligand. Configuration: $t_{2g}^6 e_g^0$. $n=0$.

Step 2: Re-evaluating the spin state

Wait, let us re-check the $Mn^{3+}$ configuration in $[Mn(CN)_{6}]^{3-}$. $Mn^{3+}$ ($3d^4$) with strong field ligand $CN^-$ results in $t_{2g}^4 e_g^0$. Unpaired electrons = 2.

Let us re-check $[Fe(CN)_{6}]^{3-}$. $Fe^{3+}$ ($3d^5$) with strong field ligand $CN^-$ results in $t_{2g}^5 e_g^0$. Unpaired electrons = 1.

Comparing $n$ values: $Ti^{3+} (n=1)$, $Mn^{3+} (n=2)$, $Fe^{3+} (n=1)$, $Co^{3+} (n=0)$.

Step 3: Calculate Magnetic Moment

The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ B.M.

For $n=2$, $\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83$ B.M., which is the highest among the given options.