= Solution
Interpret the applied <pressure> as a uniform increase of external <hydrostatic pressure> on all six faces, and approximate the rock by an <isotropic> material obeying <linear elasticity>. Taking tensile <stress> as positive, $\sigma_x=\sigma_y=\sigma_z=-p$. By <Hooke's law>, the <strain> along any one axis is
$$
\epsilon_x=\frac{\sigma_x-\nu(\sigma_y+\sigma_z)}{E}=-\frac{(1-2\nu)p}{E}.
$$
To first order in the small <strain>, the fractional <volume> change is the sum of the three axial <strains>:
$$
\frac{\Delta V}{V}=\epsilon_x+\epsilon_y+\epsilon_z=-\frac{3(1-2\nu)p}{E}=-\frac pK,
\qquad K=\frac{E}{3(1-2\nu)}.
$$
Thus the <bulk modulus> is $K=40\,\mathrm{GPa}$. Using the specified <Young's modulus>, <Poisson's ratio>, $V=1\,\mathrm{m^3}$ and $p=101325\,\mathrm{Pa}$ gives
$$
\boxed{\Delta V=-2.53\times10^{-6}\,\mathrm{m^3}=-2.53\,\mathrm{cm^3}.}
$$
The sign indicates contraction. This is the change relative to the unloaded reference state; simply maintaining a specimen already equilibrated at this same <pressure> would produce no further change.
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