In the given complex $[Cr(NH_3)_6](Br)_3$, the central chromium atom is present as $Cr^{3+}$ ion. The ground state electronic configuration of Cr (Z = 24) is $[Ar] 3d^5 4s^1$. Therefore, the electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$. In an octahedral crystal field, these three electrons will occupy the lower energy $t_{2g}$ orbitals singly ($t_{2g}^3 e_g^0$) according to Hund's rule, irrespective of the field strength of the ligand. Thus, there are 3 unpaired electrons in the chromium ion.