As the man is stationary on the belt acceleration of man=acceleration of belt
Net force on the man $=ma$
$\qquad=65 \times 1=65 N$
Coefficient of friction $\mu=0.2$
Limiting friction $=\mu mg$
For the man to be stationary with respect to belt accelerating with $'a_0'$ of belt
$\qquad=0.2 \times 9.8 $
Hence a is the correct answer.