1. For a transition metal ion, the spin-only magnetic moment is

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

where $n$ is the number of unpaired electrons.

2. Given:

$\mu = 3.86\ \text{BM}$

We check values of $n$:

• If $n=1$, then $\mu = \sqrt{3} = 1.73$
• If $n=2$, then $\mu = \sqrt{8} = 2.83$
• If $n=3$, then $\mu = \sqrt{15} = 3.87$
• If $n=4$, then $\mu = \sqrt{24} = 4.90$

So, the ion has

$n=3$

unpaired electrons.

3. We need a first-row transition metal in the $+2$ oxidation state with 3 unpaired electrons.

Let us test the options.

Option A: $Z=22$

This is Ti.

Ti: $[\text{Ar}]3d^24s^2$

$\text{Ti}^{2+} : [\text{Ar}]3d^2$

This has 2 unpaired electrons, not 3.

Option B: $Z=23$

This is V.

V: $[\text{Ar}]3d^34s^2$

$\text{V}^{2+} : [\text{Ar}]3d^3$

This has 3 unpaired electrons.

Option C: $Z=26$

This is Fe.

Fe: $[\text{Ar}]3d^64s^2$

$\text{Fe}^{2+} : [\text{Ar}]3d^6$

Free/high-spin $d^6$ has 4 unpaired electrons, not 3.

Option D: $Z=25$

This is Mn.

Mn: $[\text{Ar}]3d^54s^2$

$\text{Mn}^{2+} : [\text{Ar}]3d^5$

This has 5 unpaired electrons, not 3.

4. Therefore, the correct metal is vanadium, whose atomic number is

$\boxed{23}$

So the correct option is B.