$$P=2250 kN\space\space M20\space \&\space Fe415$$

$P_u=1.5×2250=3375 kN \\ \text { Assume 1% steel of gross c/s }\\ \text {99% concrete of gross c/s }\\ Asc =0.01 A_g\\ Ac = 0.99 A_g \\ Now \space P_{uc}=0.4 f_{ck} A_c+0.67 f_y A_{sc} \\ 3375 ×10^3=[0.4 ×20×0.99 A_g+0.67 ×415×0.01 A_g] \\ A_g=315405.82 mm^2 \\ A_g=side^2 \\ side=561.61 ≈570 mm \\ (A_g)_p=570 ×570 =324900 mm^2\\ A_{sc}=0.01×324900=3249 mm^2\\ A_{sc} p=8-25 mm∅=3928mm^2 \\ A_c p=A_{gp}-A_{scp}=324900-3928=320972 mm^2 \\ Check \\ P_u=[(0.4f_{ck} (Ac)_p )+(0.67f_y (Asc)_p)] \\ P_u=[(0.4×20×320972)+(0.67×415×3928)] \\ P_u=3659.95 kN \gt 3375kN \text { Therefore Safe.} \\ \text { Lateral Ties }\\ ∅_{ties} ≥ \dfrac {∅_{main}}4=\dfrac {25}4=6.25mm ≅8mm \\ Spacing \\ S_1=\text { least lateral Dimension }=570mm \\ S_2=16 ×∅_{main}=16×25=400mm \\ S_3=48×∅_{ties}=48×8=384mm \\ S_4=300mm \\ \text { Provided 8mm ∅ ties @ 300 mm c/c }$