Answer
The skateboard's velocity relative to the sidewalk is $-4.62m/s$
Work Step by Step
On the horizontal, at the start, the boy-skateboard system with mass $m+M=2.5+40=42.5kg$ has velocity $\vec{v_0}=+5.3m/s$
Then, the boy with mass $M=40kg$ has a horizontal velocity $V_f=+6\times\cos9.5=+5.92m/s$. The skateboard with $m=2.5kg$ has horizontal velocity $\vec{v_f}$, which is the skateboard's velocity relative to the sidewalk.
Since we neglect friction, the total linear momentum is conserved. Therefore,
$$m\vec{v_f}+M\vec{V_f}=(m+M)\vec{v_0}$$ $$\vec{v_f}=\frac{(m+M)\vec{v_0}-M\vec{V_f}}{m}$$ $$\vec{v_f}=\frac{+225.25-236.8}{2.5}=-4.62m/s$$
