= Solution
For <linear elasticity> of an isotropic material, the <isotropic elastic modulus relation> gives
$$
B=\frac{E}{3(1-2\nu)},\qquad
G_s=\frac{E}{2(1+\nu)}.
$$
The printed <Young's modulus> is $E=10^{-3}\ \mathrm{GPa}=10^6\ \mathrm{Pa}$, and the printed <Poisson's ratio> is $\nu=1.47$. Formal substitution gives
$$
B=-1.72\times10^5\ \mathrm{Pa},\qquad
G_s=2.02\times10^5\ \mathrm{Pa}.
$$
With compressive <pressure> positive, the linear volume response is $\Delta V/V=-p/B$. Thus a thirty per cent reduction formally requires
$$
\boxed{p=0.30B=-5.15\times10^4\ \mathrm{Pa}=-0.509\ \mathrm{atm},\qquad
G_s=0.202\ \mathrm{MPa}.}
$$
Here $1\ \mathrm{atm}=101325\ \mathrm{Pa}$. The negative pressure is a formal tensile load, not the requested positive isotropic compression.
The original PDF really prints $1.47$. It is incompatible with a stable ordinary isotropic elastic solid: positive <bulk modulus> and <shear modulus> require $-1<\nu<1/2$. Consequently there is no physically admissible positive-pressure answer with the stated constants. If $0.47$ was intended, the corresponding linear estimates would instead be
$$
\boxed{p\simeq1.67\times10^6\ \mathrm{Pa}=16.4\ \mathrm{atm},\qquad
G_s\simeq0.340\ \mathrm{MPa}\quad(\nu=0.47\ \text{assumed}).}
$$
That is an explicit alternative, not a correction established by the source. A thirty per cent volume strain is also outside a controlled small-strain approximation; accurate rubber behaviour at that deformation needs a finite-strain constitutive law. The initial volume cancels because the specified change is a fractional one.
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