The correct option is A 343.77
Given data Ast=π4×162×4=804.25mm2
b = 200 mm; d = 400 mm
From equilibrium of forces
T = C
Ast,fst=0.36×fck b xu,lim
⇒fst=0.36×fck b xu,limAst
=0.36×20×200×400×(0.48)804.25
∴fst=343.77N/mm2
Alternatively,
xu,lim=0.48d=0.48(400)=192mm
Mu,lim=0.138fckbd2=0.138(20)(200)(400)2Nmm
= 88.32 kNm
Mu,lim=fst,Ast(d−0.42xu)
⇒88.32×106=fst(804.25)(400−0.42×192)
⇒fst=343.86N/mm2