= Solution
Retain the same regional-background assumption as in part (b). The northern <pluton> is about $2.0\ \mathrm{km}$ thick beneath the station near the northern trough, while its <gravity anomaly> is about $-114\ \mathrm{mGal}$. Its anomaly relative to the greenstone maximum is approximately $-14\ \mathrm{mGal}$. Therefore the contrast equation is
$$
-14\times10^{-5}=2\pi G\{(\rho_{\rm gr}-2670)(2000)-(\rho_{\rm gs}-2670)(1500)\}.
$$
Using the result of part (b), the second term contributes $15\ \mathrm{mGal}$, leaving a net pluton anomaly of about $+1\ \mathrm{mGal}$. Hence \b[$\rho_{\rm gr}\simeq2680\ \mathrm{kg\,m^{-3}}$, effectively close to the host-rock <density> at the precision of this diagram].
The apparent trough is strongly negative relative to the <greenstone belt>, but hardly negative relative to the southern host-rock background. If a northward regional trend is intended instead, it must be stated: in general,
$$
\rho_{\rm gr}\simeq2670+\frac{[-114-g_{\rm bg,N}]\,10^{-5}}{2\pi G(2000)},
$$
with both <gravity> values in <milligals>. For example, a northern background of $-109\ \mathrm{mGal}$ would give $\rho_{\rm gr}\simeq2610\ \mathrm{kg\,m^{-3}}$, but the same reference would also change part (b). No regional curve is explicitly identified in the supplied figure, so the numerical result must retain its reference-field assumption. This is the geological <nonuniqueness of geophysical inversion> of potential-field interpretation, rather than a reason to force a presumed textbook <granite> density onto the plot.
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