Answer : 9.3 m/s
Step 1:
Two forces are acting on the car, gravity and the normal force.
At the top, both forces are toward the center of the circle, so Newtonâ€™s second law gives $mg+n_B=ma_{rad}$
At the bottom, gravity is downward but the normal force is upward, so $n_A-mg=ma_{rad}$
Solving for the acceleration
$a_{rad} =g+ \large\frac{n_B}{m}$
$\qquad= 9.8 m/s +\large\frac{6N}{0.8 \;kg}$
$\qquad= 17.3 \;m/s^2$
Step 2:
The speed of the car can now be calculated :
$v= \sqrt {a_{rad} R}$
$\quad= \sqrt {17.3 m/s^2.5 \;m}$
$\qquad= 9.3 m/s$