The 'spin-only' magnetic moment [in units of Bohr magneton $\left({\mu }_{\mathrm{B}}\right)$] of ${\mathrm{Ni}}^{+2}$ in aqueous solution would be: (Atomic number of $\mathrm{Ni} = 28$)

Since water is a weak ligand in aqueous solutions, high spin complexes develop.
Electrical arrangement of ${\mathrm{Ni}}_{\mathrm{i}}=\left[\mathrm{Ar}\right]4{\mathrm{s}}^{2}3{\mathrm{d}}^8$

Electrical arrangement of ${\mathrm{Ni}}^{2+}=\left[\mathrm{Ar}\right]4{\mathrm{s}}^{0}3{\mathrm{d}}^8$

${\mathrm{Ni}}^{+2},{\mathrm{d}}^8$ two unpaired electrons in the arrangement, the magnetic moment is:

$\begin{array}{l}\mu =\sqrt{n(n+2)}\\ =\sqrt{2\times 4}=2.84\mathrm{BM}\end{array}$

As a result, choice A is accurate.