Answer
$v_3=16.85m/s$
Work Step by Step
At point 1, the driver's weight and the normal force exerted on her oppose each other. We have $$F_{N_1}-mg=F_{c_1}=\frac{mv^2_1}{r}$$ $$F_{N_1}=m\Big(g+\frac{v^2_1}{r}\Big) (1)$$
At point 3, the driver's weight and the normal force exerted on her reinforce each other to give $F_c$. We have $$F_{N_3}+mg=F_{c_3}=\frac{mv^2_3}{r}$$ $$F_{N_3}=m\Big(\frac{v^2_3}{r}-g\Big) (2)$$
Therefore, for $F_{N_1}=F_{N_3}$, $$g+\frac{v^2_1}{r}=\frac{v^2_3}{r}-g$$ $$\frac{v^2_3-v^2_1}{r}=2g$$ $$v_3=\sqrt{2gr+v^2_1}$$
We have $g=9.8m/s^2$, $r=3m$ and the speed at point 1 $v_1=15m/s$ $$v_3=16.85m/s$$
