= Table mass and the strict displacement bound
The four supporting <springs> act in parallel, so their total <spring constant> is $k=4.00\times10^4\,\mathrm{N\,m^{-1}}$. The driving <angular frequency> is $\omega=2\pi(10)=62.83\,\mathrm{s^{-1}}$. The <force-driven vibration isolation> estimate gives
$$
|x_0|\simeq\frac{F_0}{m\omega^2},\qquad
\boxed{m_{\rm estimate}=\frac{5}{(2\pi\,10)^2(10^{-6})}=1.267\times10^3\,\mathrm{kg}}.
$$
At this <mass>, $f_0=\sqrt{k/m}/(2\pi)=0.894\,\mathrm{Hz}\ll10\,\mathrm{Hz}$, and $Q=\sqrt{mk}/b\simeq14.2$, so the assumptions are consistent. The exact forced amplitude is
$$
|x_0|=\frac5{\sqrt{(4.00\times10^4-m\omega^2)^2+(500\omega)^2}}
=\boxed{1.008\,\mathrm{\mu m}}.
$$
It is indeed close to $1\,\mathrm{\mu m}$, but it is slightly \b[above] the strict requested bound. Setting the exact expression to $10^{-6}\,\mathrm m$ gives the high-mass threshold
$$
\boxed{m_{\min}=\frac{k+\sqrt{(F_0/x_{\max})^2-b^2\omega^2}}{\omega^2}=1.277\times10^3\,\mathrm{kg}}.
$$
Choose a <mass> strictly greater than this threshold for a strict inequality; $1.28\times10^3\,\mathrm{kg}$ is a suitable rounded choice. Simply rounding the high-frequency estimate down would not guarantee the bound.
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