= Solution
For a wide horizontal <Bouguer slab> replacing surrounding basement, use the \b[density contrast] $\Delta\rho=\rho_{\mathrm{sed}}-\rho_{\mathrm{base}}$, not the absolute <sediment> <density>:
$$
\Delta g=2\pi G\Delta\rho\,h.
$$
Take downward attraction as positive. The negative anomaly therefore requires $\Delta\rho<0$. With $\Delta g=-40\ \mathrm{mGal}=-4.0\times10^{-4}\ \mathrm{m\,s^{-2}}$, $h=3000\ \mathrm m$, and $G=6.67430\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}$,
$$
\Delta\rho=\frac{-4.0\times10^{-4}}{2\pi(6.67430\times10^{-11})(3000)}=-317.95\ \mathrm{kg\,m^{-3}},
$$
so
$$
\boxed{\rho_{\mathrm{base}}=2300-(-317.95)\simeq2618\ \mathrm{kg\,m^{-3}}\simeq2.62\ \mathrm{Mg\,m^{-3}}.}
$$
This is the requested <Bouguer slab> approximation: a flat observation surface, laterally infinite constant-thickness material, uniform average <geophysical density contrast>, and a regional anomaly already removed. The actual basin in Figure 1 narrows downwards and is finite. For a fixed maximum thickness, finite width reduces the central anomaly magnitude, so this calculation can underestimate the true magnitude of the average <geophysical density contrast>. It does not establish a unique basement <density> without the geometrical assumption.
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