The helical reinforcement consists of $8mm$ clear cover to main steel equal to $50mm$ Use $M20/Fe415.$

For helical reinforcement.

$$\dfrac {\pi×D_k×a_{sh}}{pitch \space A_k }\space \&\space \dfrac {0.36×f_{ck}}{/f_y} ×(\dfrac {A_q}{A_k} -1) $$

If LHS ≥ RHS than {P_u} will be

$$P_u=1.05×[(0.4f_{ck} A_c )+(0.4f_y A_{sc})]$$

If RH > LHS than $P_u$ will be

$$P_u=[(0.4f_{ck} A_c )+(0.67f_y A_{sc})]$$

Diameter of column $=350mm.$

$$Asc = 6-20mm∅ = 1884mm^2 $$ $ A_q=\dfrac \pi4 (350)^2=96211mm^2\\ A_c= A_q- A_{sc}=94327mm^2 \\ Dk=D-(2 ×clear cover)=350-(2×50)=250 \\ A_k=\dfrac \pi4×250^2=49087mm^2 \\ a_{sh}=\dfrac \pi4×8^2=50 mm^2\\ Pitch = 50mm \\ f_{ck}=20N/mm^2 f_y=415 N/mm^2 \\ \dfrac {\pi×250×50}{50×49087} \space \&\space \dfrac {0.36×20}{415} ×(\dfrac {96211}{49087}-1) \\ 16 ×10^{-3}\space \&\space 16.65 ×10^{-3}\\ \text { Therefore RHS \gt LHS }\\ P_u \text { will not contain }1.05 \\ P_u=[(0.4f_{ck} A_c )+(0.62f_y A_{sc})] \\ P_u=[(0.4×20×94327)+(0.62×415×1884)] \\ P_u=1239.37 kN$