Data:- $b =1200 mm \\ D = 120 mm \\ b =230 mm \\ D = 6000 mm \\ Ast = 4-22 mm =1520.53mm^2 \\ M20, F_{e415} \\ D= D – dc = 550 mm \\ \text {Assume X_u \lt D_f [N.A lies in the flange]} \\ C_u=T_u\\ 0.36 f_ckb_fX_u=0.87 f_y Ast \\ 0.36\times20\times1200\times X_u=0.87\times415\times1520.33\\ X_u=63.54 mm \lt D_f $

Assumption is correct

Now, $X_{u\space max} = 0.48d = 0.48 x 550 = 264mm$

$X_u \lt X_{u\space max} …..$ Hence under reinforced section

$M_u=(C_(u1)\times L_{a1}) \\ M_u=[(0.36f_ckb_wX_{u\space max })\times (d-0.42X_{u\space max})]\\ M_u=0.36\times 20\times1200\times63.59\times(550-0.42\times63.54)\\ M_u=287.29 KNm$