Correct Answer - C
In case of `d^(3)` configuration, the number of unpaired electrons remains 3 whether the ligand is strong field or weak field. The hybridisation scheme can be shown as follow.
Hence the complex is inner orbital complex as it involves `(n-1) d` orbitals for hybridisation , `3.93 = sqrt(n(n+2))` , so `n=3` ( here n is numbr of unpaired electron (s)).