1. Idea: For octahedral species, two kinds of hybridization are commonly discussed:

   • $d^2sp^3$: inner orbital complex
   • $sp^3d^2$: outer orbital complex

   We must count how many of the given species are outer orbital octahedral complexes, i.e. involve $sp^3d^2$ hybridization.

2. Analyze each species one by one.

(i) $[Co(NH_3)_6]^{3+}$

• Oxidation state of Co: $x+6(0)=+3 \Rightarrow x=+3$
• So, $Co^{3+}$ has configuration: $Co: [Ar]3d^74s^2 \,\Rightarrow\, Co^{3+}: 3d^6$
• $NH_3$ is a sufficiently strong ligand for $Co^{3+}$, causing pairing.
• Hence it forms an inner orbital octahedral complex: $d^2sp^3$
• Not $sp^3d^2$.

(ii) $SF_6$

• Sulfur forms six bonds with fluorine in an octahedral arrangement.
• In classical hybridization language, this is taken as: $sp^3d^2$
• So this is involved in $sp^3d^2$ hybridization.

(iii) $[CrF_6]^{3-}$

• Oxidation state of Cr: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $Cr^{3+}$ has configuration: $Cr: [Ar]3d^54s^1 \,\Rightarrow\, Cr^{3+}: 3d^3$
• For octahedral $d^3$, two 3d orbitals are empty already.
• Hence it uses inner orbitals: $d^2sp^3$
• Not $sp^3d^2$.

(iv) $[CoF_6]^{3-}$

• Oxidation state of Co: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $Co^{3+}=3d^6$.
• $F^-$ is a weak field ligand, so electrons do not pair up.
• Therefore it forms an outer orbital octahedral complex: $sp^3d^2$
• So this is involved in $sp^3d^2$ hybridization.

(v) $[Mn(CN)_6]^{3-}$

• Oxidation state of Mn: $x+6(-1)=-3 \Rightarrow x=+3$
• So, $Mn^{3+}$ has configuration: $Mn: [Ar]3d^54s^2 \,\Rightarrow\, Mn^{3+}: 3d^4$
• $CN^-$ is a strong field ligand, so pairing occurs.
• Thus it forms an inner orbital complex: $d^2sp^3$
• Not $sp^3d^2$.

(vi) $[MnCl_6]^{3-}$

• Oxidation state of Mn is again $+3$, so $Mn^{3+}=3d^4$.
• $Cl^-$ is a weak field ligand, so no pairing occurs.
• Therefore it forms an outer orbital octahedral complex: $sp^3d^2$
• So this is involved in $sp^3d^2$ hybridization.

3. Count the species with $sp^3d^2$ hybridization:

• $SF_6$
• $[CoF_6]^{3-}$
• $[MnCl_6]^{3-}$

So total number = $3$

4. Correct option: $\boxed{B}$