Answer
The speed at the bottom of the ramp is $6.57m/s$
Work Step by Step
As we can see in the image below, since we neglect friction, the only force acting on the student's motion down the ramp is $mg\sin\theta$, where here $\theta=18^o$. Therefore, $$\sum F=m_{stu}g\sin18^o=3.03m_{stu}N$$
From Newton's 2nd Law, the student's acceleration can be calculated: $$a=\frac{\sum F}{m_{stu}}=3.03m/s^2$$
We have $v_0=2.6m/s$, $a=3.03m/s^2$ and the ramp's length $x=6m$. The speed at the bottom of the ramp would be $$v^2=v_0^2+2ax=43.12m^2/s^2$$ $$v=6.57m/s$$
