= Solution
For fixed atom number, volume scales as the cube of the touching-sphere radius. Compressing silver to the stated gold lattice size therefore requires
$$
\frac{V_f}{V_i}=\left(\frac{1.442}{1.444}\right)^3=0.9958506,
$$
a volume decrease of about $0.415\%$. If the <bulk modulus> is constant,
$$
\mathrm dp=-B\frac{\mathrm dV}{V},\qquad
\Delta p=B\ln\frac{V_i}{V_f}
=3B\ln\frac{1.444}{1.442}.
$$
Using the silver value derived above gives
$$
\boxed{\Delta p=0.1436\ \mathrm{GPa}\simeq1.44\times10^8\ \mathrm{Pa}.}
$$
For this small compression, $\Delta p\simeq-B\Delta V/V$ gives almost the same estimate. This is an additional pressure relative to the initial standard-pressure state, using the question's free-electron model modulus rather than an empirical silver modulus.
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