Answer
$$\frac{F_N}{W}=8.82$$
Work Step by Step
At the bottom of a dive, the normal force and the weight cancel each other out to give the centripetal force $F_c$. $$F_N-W=F_c$$
Divide both sides by $W$: $$\frac{F_N}{W}-1=\frac{F_c}{W}=\frac{\frac{mv^2}{r}}{mg}=\frac{v^2}{rg}$$ $$\frac{F_N}{W}=\frac{v^2}{rg}+1$$
We have $v=230m/s$, $r=690m$ and $g=9.8m/s^2$. So,
$$\frac{F_N}{W}=8.82$$
