Answer
$2.431\;\rm km$
Work Step by Step
According to table 13.1, the free-fall acceleration at sea level is given by
$$g_{\rm ground}=\bf9.83\;\rm m/s^2$$
where
$$g_{\rm ground}=\dfrac{GM_E}{R_E^2}=\bf 9.83\;\rm m/s^2\tag 1$$
And our sensitive gravimeter at a mountain observatory finds that
the free-fall acceleration is
$$g_{\rm mountain}=9.83-0.0075 =\bf 9.8225\;\rm m/s^2$$
where
$$g_{\rm mountain}=\dfrac{GM_E}{(R_E+h)^2}=\bf 9.8225\;\rm m/s^2\tag 2$$
Now we need to find $h$ which is the height of the mountain.
Solving (1) for $GM_E$ and then plug it into (2);
$$\dfrac{9.83R_E^2}{(R_E+h)^2}= 9.8225 $$
$$ 9.83R_E^2 = 9.8225(R_E+h)^2 $$
$$\dfrac{ 9.83}{9.8225}R_E^2 = (R_E+h)^2 $$
Taking the square toot for both sides;
$$ \sqrt{\dfrac{9.83}{9.8225}}R_E = R_E+h $$
Thus,
$$h=\sqrt{\dfrac{9.83}{9.8225}}R_E - R_E=R_E\left[\sqrt{\dfrac{9.83}{9.8225}}-1\right]$$
Plugging $R_E$ from table 13.2;
$$h =(6.37\times 10^6)\left[\sqrt{\dfrac{9.83}{9.8225}}-1\right]$$
$$h=\color{red}{\bf 2.431\times 10^3}\;\rm m$$