For the 8 kg block (C) to move with constant velocity, Fnet=0.
Friction forces acting:
Between C and ground: f=μ(mA+mB+mC)g=0.5×(4+6+8)×10=90 N
Between C and B: f=μ(mA+mB)g=0.5×(4+6)×10=50 N
For the 6 kg block (B):
T=fB on ground+fA on B
T=μ(mA+mB)g⋅mA+mBmB+μ⋅mA⋅g
T=20+50=70 N
For the 8 kg block (C):
F=90+T+50=90+70+50=210 N