b
Let $a$ be the maximum acceleration of man.
The block will remain fixed if horizontal component of tension in the rope is equal to the friction force of block against the ground. i.e.
$T \cos 37^{\circ}=f=\mu\left(m_{B} g-T \sin 37^{\circ}\right)$
$\frac{4 T}{5}=\frac{100 g}{3}-\frac{T}{5}$
$\therefore T=\frac{100 g}{3}$
Now, $m_{A} a=T-m_{A} g$
or, $25 a=\frac{100}{3} g-25 g$
or, $a=\frac{4}{3} g-g=\frac{g}{3}$
