= Solution
In the classical equilibrium circuit, the <capacitor> stores $CV^2/2$. <Equipartition> assigns this quadratic term mean <energy> $k_BT/2$, so
$$
\boxed{V_{\rm rms}=\sqrt{\frac{k_BT}{C}}\simeq6.4\,\mathrm{\mu V}\quad(T\simeq300\,\mathrm K).}
$$
The mean <voltage> is zero and its <variance> is $k_BT/C$. Resistance determines the equilibration time and noise spectrum, not this full-band equilibrium <variance>. A finite-bandwidth or appreciably loading readout can measure less than the ideal result.
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