Width of flange = $800$ mm

Depth of slab = $80$ mm

Width of rub = $300$ mm

Area of steel = $4-20$ mm on tension side

D = $450$ mm, dc = $50$ mm

Data:- $b_f =800 mm \\ D_f = 80 mm \\ b_w =300 mm \\ d = 450 mm \\ Ast = 4-20 mm \\ =1256.63 mm^2\\ M20, F_{e415}$

Assume $X_u \lt D_f$ [N.A lies in the flange]

$C_u=T_u\\ 0.36f_ckb_fX_u=0.87f_yAst\\ 0.36\times 20\times800\times X_u=0.87\times415\times1256.63 \\ X_u=78.76mm \lt D_f $

Assumption is correct

Now $X_{u\space max}=0.48d=0.48\times 450=216mm\\ X_u \lt X_{u\space max} ... \text{ Hence under reinforced section. }\\ M_u=(C_{u1}\times L_{a1})\\ M_u=[(0.36f_ckb_wX_{u})\times (d-0.42 X_{u})]\\ M_u =0.36\times20\times800\times78.76\times(4550-0.42\times78.76) \\ M_u=189.13 KNm$

The answer is correct, but you need to correct the formula of MU. In under reinforced section Xu is used in Formula and You wrote Xumax, and after writing Xumax, you took the Xu value..

Xu = 78.76mm XuMax = 216mm