Answer
$$0.36 \mathrm{kg} / \mathrm{yr}$$
Work Step by Step
Given $$W(t)=\left(0.14+0.115 t-0.002 t^{2}+0.000023 t^{3}\right)^{3.4}$$
Since
$$W'(t)= 3.4\left(0.14+0.115 t-0.002 t^{2}+0.000023 t^{3}\right)^{2.4}\left(0.115-0.004 t+0.000069 t^{2}\right) $$
Then
\begin{align*}
W'(10)&= 3.4\left(0.14+0.115(10)-0.002 (10)^{2}+0.000023 (10)^{3}\right)^{2.4}\left(0.115-0.004 (10)+0.000069 (10)^{2}\right)\\
&\approx 0.36 \mathrm{kg} / \mathrm{yr}
\end{align*}
