Given $v$ is uniformly distributed, Therefore P.D.F of $v$ is-
$$ \ F(v) = \left\{ \begin{array}{ll} \frac{1}{10-0} & 0\leq v\leq10 \\ \\ 0 & otherwise \end{array} \right. $$
Also $W(v) $= 0.003$v^{2}$,
$ \ E[w(v)] = \int_{0}^{10} W(v)F(v)dv \\ \\ E[w(v)] = \int_{0}^{10} 0.003v^{2} * \frac{1}{10} dv$
solving this we get $E[w(v)] = 0.1$
Reference: Google Books