So basically the bike has to travel 3.9 meters, with that in mind it's a typical projectile motion problem with the only relevant acceleration coming from gravity, so you can use your kinematic equations
Remember that if the bike leaves the ramp at speed Vi, then Vix=V*cos(theta) and Viy=V*sin(theta) where the Vix and y are the x and y components of V. If you don't wanna remember this and can't derive it from basic trig, remember that if the ramp were flat(theta=) ALL the velocity would be in the x direction, ie V=Vix, and since cos(0)=1, that's the trig function that makes it true(and conversely if the ramp were straight up at theta=90, V=Viy and sin(90)=1)
Remember that the distance it travels in the x(horizontal) direction is just d=Vix*t, or to be more useful for us, Vix=d/t, we know d=3.9m but we need t, which we get from looking at the y direction
You could use d=Viy*t-1/2*g*t^2, but what's d? It's 0, because the bike takes off and lands from the same height, in other words d=yfinal-yinitial=0
So 0=Viy*t-g/2*t^2, this is a very easy quadratic equation because you can just factor out t and divide both sides by t(because actually one solution to the quadratic is just t=0, because the bike has gone d=0 when it hasn't even moved yet, and it's gone d=0 after it's taken off and landed, which is the answer we want) to get 0=Viy-g/2*t
t=2*Viy/g, write as t=2*V*sin(theta)/g, and write that first equation as t=d/(V*cos(theta)) and set them equal to each other(t=t) and rearrange using algebra
2*V^2*sin(theta)*cos(theta)/g=d
V=sqrt(g*d/(2*sin(theta)*cos(theta))), we know theta=15degrees, d=3.9m, and g=9.8m/s^2 so plug that in for V=8.745 m/s
