1. Identify the oxidation state and geometry

Given the complex as $[M(H_2O)_6]Cl_2$, the complex cation is $[M(H_2O)_6]^{2+}$.

• Since $H_2O$ is a neutral ligand, the oxidation state of $M$ is $+2$.
• The complex is octahedral with six water ligands.
• $H_2O$ is a weak-field ligand, so for first-row transition metals this usually gives high-spin complexes.

2. Use the spin-only magnetic moment formula

The spin-only magnetic moment is:

$\mu = \sqrt{n(n+2)}\; \text{BM}$

where $n$ is the number of unpaired electrons.

We are given:

$\mu \approx 3.9\; \text{BM}$

Now check values:

• For $n=1$: $\mu = \sqrt{3} \approx 1.73$
• For $n=2$: $\mu = \sqrt{8} \approx 2.83$
• For $n=3$: $\mu = \sqrt{15} \approx 3.87$
• For $n=4$: $\mu = \sqrt{24} \approx 4.90$
• For $n=5$: $\mu = \sqrt{35} \approx 5.92$

So $3.9$ BM corresponds to $n=3$ unpaired electrons.

3. Find which $M^{2+}$ ions have 3 unpaired electrons in octahedral high-spin complexes

Let's write the $d$-electron counts:

• $V^{2+}$: $3d^3$
• $Cr^{2+}$: $3d^4$
• $Mn^{2+}$: $3d^5$
• $Fe^{2+}$: $3d^6$
• $Co^{2+}$: $3d^7$

In octahedral high-spin cases:

• $d^3 \to t_{2g}^3 e_g^0$ : 3 unpaired
• $d^4 \to t_{2g}^3 e_g^1$ : 4 unpaired
• $d^5 \to t_{2g}^3 e_g^2$ : 5 unpaired
• $d^6 \to t_{2g}^4 e_g^2$ : 4 unpaired
• $d^7 \to t_{2g}^5 e_g^2$ : 3 unpaired

Thus the ions giving $3$ unpaired electrons are:

$V^{2+} \text{ and } Co^{2+}$

4. Check the options

• A: $V^{2+}$ and $Fe^{2+}$ $\to$ $3$ and $4$ unpaired, so incorrect
• B: $V^{2+}$ and $Co^{2+}$ $\to$ both have $3$ unpaired, so correct
• C: $Co^{2+}$ and $Fe^{2+}$ $\to$ $3$ and $4$ unpaired, so incorrect
• D: $Cr^{2+}$ and $Mn^{2+}$ $\to$ $4$ and $5$ unpaired, so incorrect

5. Final answer

The correct pair is:

$\boxed{V^{2+} \text{ and } Co^{2+}}$

So, Option B is correct.