The context states the general order of filling as 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s... After 3p, 4s is filled next. This is also consistent with the (n+l) rule where (n+l) for 3p is (3+1)=4 and for 4s is (4+0)=4. When (n+l) values are equal, the orbital with the lower 'n' value has lower energy. However, in the given sequence, 4s (n=4, l=0) has higher n but is filled before 3d (n=3, l=2). The general order provided in the NCERT text specifies 4s before 3d, making 4s the correct answer. (Reference: 'However, following order of energies of the orbitals is extremely useful: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s...')