Considering $H_2O$ as a weak field ligand, the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ will be – (At. no. of Mn = 25)

Electronic configuration of Mn (Z = 25): $[Ar]\,3d^5\,4s^2$
In $[Mn(H_2O)_6]^{2+}$, Mn is in the +2 state: $Mn^{2+} = [Ar]\,3d^5$
$H_2O$ is a weak field ligand, so the crystal field splitting $\Delta_o$ is smaller than the pairing energy. Electrons occupy the $t_{2g}$ and $e_g$ orbitals singly before pairing (high spin).
Configuration: $t_{2g}^3\,e_g^2$, with all five electrons unpaired.
So the number of unpaired electrons is five (magnetic moment about 5.92 BM).