The electronic configuration of the last three orbitals in cerium (At. No-57) in ground state is:

$4f^15d^16s^2$

The correct answer is option 1. $4f^15d^16s^2$.

The electronic configuration of cerium (Ce), with atomic number 57, can be determined by following the Aufbau principle, Hund's rule, and the Pauli exclusion principle.

The electronic configuration of cerium (Ce) in ground state is:

$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^1 5d^1 6s^2$

The last three orbitals are $4f^1, 5d^1, 6s^2$.