For a factored bending moment $1×10^8 Nmm$ Use $M_{20}/ F_{e415}$ by ultimate load theory.

Assume R.C. beam $300m × 450mm$ (effective)

b=300mm.

d = 450mm.

$$σ_{cu}=f_{ck}=20N/mm^2 \\ σ_{sy}=f_y=415 N/mm^2 \\ M_u = 1×10^8 Nmm=100 kNm $$ To find = Ast =?

We know $M_{umax}=0.25 f_{ck}bd^2=0.25×20×300×450^2=303.75kNm$ $$M_u\ltM_{umax}$$

Hence it is singly reinforced section.

$$M_u= C_u×l_a=\frac{2}{3}×f_ck b×a×\Big(d-\frac{a}{2}\Big)\\ 100 ×10^6=2/3×20×300×a×\Big(450-\frac{a}{2}\Big)$$ a = 194.24mm $$M_u= T_u×l_a=f_y×Ast×\Big(d-\frac{a}{2}\Big) \\ 400 ×10^6=415×Ast×\Big(450-\frac{194.24}{2}\Big) \\ Ast = 682.84 mm^2$$ $Provide 4-16mm∅ \\ Astp = 804.24 mm^2$