Answer
$N=6.18KN$
Work Step by Step
The required normal force can be determined as
$a_n=\frac{v^2}{\rho}$
$\implies v=\sqrt{\rho a_n}=\sqrt{(800)(78.5)}=251m/s$
Now, $\Sigma F_n=ma_n$
$\implies N-W=ma_n$
$\implies N=m(g+a_n)=m(g+8g)=9mg$
We plug in the known values to obtain:
$N=9(70)(9.81)$
$\implies N=6.18KN$