The tension reinforcement consists of $4-16mm∅$ .Take $σ_{cu}=20N/mm^2 \ and \ σ_{sy}=420N/mm^2$.Use ULM.

Given: b=300mm

d = 450mm.

$$σ_{cu}=f_{ck}=20N/mm^2 \\ σ_{sy}=f_y=420N/mm^2$$

$Ast = 4-16 mm∅ = 4×\frac{Π}{4}×16^2=804 mm^2$

To find $M_u= ?$

To find depth of actual N.A.

$$C_u=T_u \\ \frac{2}{3} f_{ck} ba= f_y Ast \\ \frac{2}{3}×20×300×a= 420×804$$

a=84.44mm

$a_{max}=\frac{d}{2}=\frac{450}{2}=225mm$

Here $a \lt a_{max}→$ Under reinforced section

$$M_u= T_u×l_a=f_y×Ast×\Big(d-\frac{a}{2}\Big) \\ M_u=420×804×\Big(450-\frac{84.44}{2}\Big) \\ M_u=137.7 kNm$$