I don't think the provided solution is correct because it's not dimensionally consistent. The quantity [tex]8h-1[/tex] in particular doesn't make sense since it's mixing a distance with a dimensionless constant.
Here's how I think the proper answer should look:
The net force on the box acting perpendicular to the ramp is
[tex]\sum F_\perp = F_{\rm normal} - mg \cos(\theta) = 0[/tex]
where [tex]F_{\rm normal}[/tex] is the magnitude of the normal force due to contact with the ramp and [tex]mg\cos(\theta)[/tex] is the magnitude of the box's weight acting in this direction. The net force is zero since the box doesn't move up or down relative to the plane of motion.
The net force acting parallel the ramp is
[tex]\sum F_\| = mg\sin(\theta) - F_{\rm friction} = ma[/tex]
where [tex]mg\sin(\theta)[/tex] is the magnitude of the parallel component of the box's weight, [tex]F_{\rm friction}[/tex] is the magnitude of kinetic friction, and [tex]a[/tex] is the acceleration of the box.
From the first equation, we find
[tex]F_{\rm normal} = mg \cos(\theta)[/tex]
and since [tex]F_{\rm friction} = \mu F_{\rm normal}[/tex], we get from the second equation
[tex]mg\sin(\theta) - \mu mg\cos(\theta) = ma[/tex]
and with [tex]\mu = 0.25[/tex] and [tex]\theta=60^\circ[/tex], we get
[tex]a = g\sin(60^\circ) - 0.25g \cos(60^\circ) = \left(\dfrac{\sqrt3}2 - \dfrac18\right) g[/tex]
Let [tex]x[/tex] be the length of the ramp, i.e. the distance that the box covers as it slides down it. Then the box attains a final velocity [tex]v[/tex] such that
[tex]v^2 = 2ax[/tex]
From the diagram, we see that
[tex]\sin(\theta) = \dfrac hx \implies x = \dfrac h{\sin(60^\circ)} = \dfrac{2h}{\sqrt3}[/tex]
and so
[tex]v^2 = \dfrac{4ah}{\sqrt3} = \left(2 - \dfrac1{2\sqrt3}\right) hg = \dfrac14 \left(8 - \dfrac2{\sqrt3}\right) hg[/tex]
[tex]\implies v = \dfrac12 \sqrt{\left(8 - \dfrac2{\sqrt3}\right) hg}[/tex]