The value of the 'spin only' magnetic moment for one of the following configurations is $$2.84$$ BM. The correct one is

The spin-only magnetic moment of a transition-metal ion (no orbital contribution) is obtained from

$$\mu_{\text{spin}} = \sqrt{n(n+2)}\;{\text{BM}}$$
where $$n$$ is the number of unpaired electrons.

The observed value is $$\mu_{\text{obs}} = 2.84\;\text{BM}$$.
Set $$\sqrt{n(n+2)} = 2.84$$ and test small integers:

$$n = 1 \;\Rightarrow\; \sqrt{1(1+2)} = \sqrt3 = 1.73\;\text{BM}$$

$$n = 2 \;\Rightarrow\; \sqrt{2(2+2)} = \sqrt8 = 2.83\;\text{BM}$$

The calculated value for $$n = 2$$ (2.83 BM) matches the given 2.84 BM, so the ion must contain exactly two unpaired electrons.

Now examine each electronic configuration provided:

Crystal-field splitting is $$t_{2g}^{\;6}$$ > $$e_g^{\;4}$$, and pairing is favoured.
Electron filling: $$t_{2g}^{\,4}\;e_g^{\,0}$$
Arrangement inside $$t_{2g}$$ orbitals: one orbital is doubly occupied, the other two are singly occupied → $$n = 2$$ unpaired.

Electron filling: $$t_{2g}^{\,3}\;e_g^{\,1}$$ → three unpaired in $$t_{2g}$$ + one unpaired in $$e_g$$ ⇒ $$n = 4$$, giving $$\mu = 4.90\;\text{BM}$$, not 2.84 BM.

Filling: $$t_{2g}^{\,3}\;e_g^{\,0}$$, all electrons remain unpaired → $$n = 3$$, so $$\mu = 3.87\;\text{BM}$$.

Filling: $$t_{2g}^{\,5}\;e_g^{\,0}$$. Four electrons pair and one remains unpaired → $$n = 1$$, giving $$\mu = 1.73\;\text{BM}$$.

Only Case 1 supplies exactly two unpaired electrons and therefore the required magnetic moment.

Hence the configuration that gives $$2.84\;\text{BM}$$ is

Option A which is: $$d^{4}$$ (in strong ligand field)