The correct answer is (b) 216.49 kN
Easiest explanation: fu = 410 MPa, fy = 250 MPa, γm0 = 1.1, γm1 = 1.25
Avg = ( 1×100 + 50 ) x 8 = 1200 mm^2
Avn = (1×100 + 50 – (2 – 1/2)x20 ) x 8 = 1440 mm^2
Atg = 35 x 8 = 280 mm^2
Atn = (35 – 1/2 x 20) x 8 = 200 mm^2
Tdb1 = (Avgfy/√3 γm0)+(0.9Atnfu/γm1) = [(1200×250/ 1.1x√3) + (0.9x200x410 / 1.25) ] x 10-3 = 216.49 kN
Tdb2 = (Atgfy/ γm0)+(0.9Avnfu/√3 γm1) = [ (0.9x1440x410 / 1.25x√3) + (280×250/1.1) ] x 10-3 = 309.06 kN
Block shear strength of tension member is 216.49 kN.