The electrons are filled in the increasing order of (n+l) value.
- 4+1=5
- 4+0=4
- 3+2=5
- 3+1=4
If (n+l) value is same for two different n and l value, then electron first fills in the orbital in which the n value is less.
The correct order is 4<2<3<1.
Hence, option B is correct.