= Solution
Treat the quoted anomaly as the response of a slab of thickness $h$ replacing host rock. The <Bouguer slab> formula applies to the <geophysical density contrast>, not the absolute density:
$$
\Delta g=2\pi G(\rho_b-\rho_h)h.
$$
With $\Delta g=-2.5\times10^{-5}\,\mathrm{m\,s^{-2}}$, $h=400$ m and $\rho_h=2750\,\mathrm{kg\,m^{-3}}$,
$$
\rho_b=2750+\frac{-2.5\times10^{-5}}{2\pi(6.67430\times10^{-11})(400)}
=2750-149.04.
$$
Thus \b[$\rho_b\simeq2601\,\mathrm{kg\,m^{-3}}$] in the slab approximation. The negative anomaly requires less dense infill. A finite bowl has a different geometric response, and water, post-impact <sediment> and the regional baseline can matter; the stated calculation is the requested slab estimate. The plotted minimum appears closer to $-2$ mGal, but the numerical instruction explicitly supplies $-2.5$ mGal and controls this calculation.
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