= Solution
\b[The volume alone does not determine a unique change of elevation.] A <local isostatic response to an added surface layer> depends on its thickness and <density>, so the <Deccan Traps>' footprint must be specified or retained as a parameter.
Let the deposited <basalt> have uniform thickness $H=V/A$ and <density> $\rho_b$, where $V=10^{15}\,\mathrm{m^3}$ and $A$ is the loaded area. Let $w$ be downward displacement of the original land surface, and let $\rho_m$ be the <density> of displaced <mantle>. Assume a subaerial, laterally extensive deposit, negligible air <density>, unchanged thickness and <density> of the pre-existing <crust>, and complete local <isostatic adjustment>. Take the <magma> supply to be external to the local compensation column, and neglect cooling, <erosion>, <magmatic underplating> and any deep thermal or source-depletion anomaly.
Choose a <compensation depth> beneath all the displaced interfaces. The added column <mass> per unit area is $\rho_b H$. Downward translation of the pre-existing <crust> replaces a thickness $w$ of <mantle> within the column, reducing its <mass> by $\rho_m w$. Equality of <hydrostatic pressure> at the <compensation depth> therefore gives
$$
\rho_b gH-\rho_m gw=0,
\qquad
\boxed{w=\frac{\rho_b}{\rho_m}\frac{V}{A}}.
$$
The newly formed top of the <basalt> stands $H$ above the depressed original surface. Consequently the change of the region's \i[top-surface] elevation is
$$
\boxed{\Delta z=H-w=\left(1-\frac{\rho_b}{\rho_m}\right)\frac{10^{15}}{A}\ \mathrm{m},\quad A\text{ in }\mathrm{m^2}}.
$$
Thus the original substrate \b[subsides], but the top of the deposit \b[rises] if $\rho_b<\rho_m$. Distinguishing these two surfaces prevents reporting the large downward displacement as the change in plateau elevation.
For an explicitly illustrative choice $A=10^{12}\,\mathrm{m^2}$, $\rho_b=2900\,\mathrm{kg\,m^{-3}}$ and $\rho_m=3300\,\mathrm{kg\,m^{-3}}$, the deposit is 1000 m thick, the substrate subsides about 879 m, and \b[the final surface rises about 121 m]. This is a conditional estimate, not a number fixed by the paper. Half that footprint would double all three heights under the same assumptions. For a nonuniform deposit, local <isostasy> gives the same relation pointwise with $H(x,y)$; the total volume still cannot specify a unique local elevation.
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