A Rh+ve man (heterozygous) is married to a Rh-ve lady. They had four children. Only first and the last are Rh-ve. Then the child with Erythroblastosis foetalis / HDNB among the four children probably is      

Male is heterozygous for Rh+ve (H,h) and female is homozygous for Rh-ve (h,h).

Male can give two types of gametes with either H or h allele for Rh and female will give only one type of gamete with h allele. So the offspring will be either Rh+ve (heterozygous) or Rh-ve.

If first and last of the four children are Rh-ve, then 2nd and 3rd child will be Rh+ve.

3rd child will suffer from erythroblastosis foetalis as mother get sensitised after the birth of 2nd child.