Answer
$$-1.15 \times 10^{-5} \mathrm{kPa} / \mathrm{m}$$
Work Step by Step
Given $$
T=15.04-0.000649 h, \quad P=101.29+\left(\frac{T+273.1}{288.08}\right)^{5.256}
$$
Since for $h=3000$, we get $$ T(3000)= 13.093$$
and
\begin{align*}
\frac{d P}{d h}&=\frac{d P}{d T} \frac{d T}{d h}\\
&= 5.256\left(\frac{T+273.1}{288.08}\right)^{4.256}\left(\frac{1}{288.08}\right) (-0.000649)
\end{align*}
Then, at $h=3000$
\begin{align*}
\frac{d P}{d h}\bigg|_{h=3000} &= 5.256\left(\frac{(13.093)+273.1}{288.08}\right)^{4.256}\left(\frac{1}{288.08}\right) (-0.000649) \\
&\approx -1.15 \times 10^{-5} \mathrm{kPa} / \mathrm{m}
\end{align*}
