Calculate ultimate load and ultimate moment the column can resist if it is just on the verge of cracking. Take $ku =1$ Use $M20/Fe415$ take $fs_1=355N/mm^2 ,fs_2=70N/mm^2 fc_1=9N/mm^2 ,fc_2=3N/mm^2$

Now

$$P_{uc}=0.36 f_{ck} bd=0.36×20×300×500=1080 kN $$

$P_{us1}=(fsc_1- fcc_1)×Asc \\ =(355-9)×3000 \\ P_{us1}=1038 kN \\ P_{us2}=(fsc_2- fcc_2 )×Asc_2=(70-3)×3000=201kN \\ Total\space P_u= P_{us1}+P_{us2}=1038+201=2319kN \\ M_{uc}=P_{uc}×(0.5-0.42)×D \\ =1080×(0.08)×500 \\ =43.2kNm \\ M_{us1}=P_{us1}×e_1=(1038×0.2)=207.6kNm \\ M_{us1}=P_{us2}×e_2=-(201×0.2)=-40.2kNm \\ M_u=167.4 kNm $