= Solution
Let $E$ denote the fluctuating microscopic <energy> and $U=\langle E\rangle$ its mean, at fixed $N,V$. In a <canonical ensemble> with $\beta=1/(k_BT)$,
$$
U=-\partial_\beta\ln Z,\qquad
\partial_\beta^2\ln Z=\langle E^2\rangle-\langle E\rangle^2.
$$
Since $dT/d\beta=-k_BT^2$, this gives
$$
\boxed{\langle(\Delta U)^2\rangle=-\partial_\beta U
=k_BT^2\left(\frac{\partial U}{\partial T}\right)_{V,N}
=k_BT^2C_V}.
$$
Here $\Delta U$ denotes an <energy> fluctuation about the mean, as in the question. The relation is not restricted to ideal gases when the <Hamiltonian operator> has no explicit <temperature> dependence. Alternatively, the stated equilibrium free-energy curvature gives a Gaussian <energy> weight proportional to $\exp[-(\Delta U)^2/(2k_BT^2C_V)]$, with the same <variance>.
For the <relative energy fluctuation of a monatomic ideal gas>, $U=3Nk_BT/2$ and $C_V=3Nk_B/2$, hence
$$
\frac{\sqrt{\langle(\Delta U)^2\rangle}}{U}
=\boxed{\sqrt{\frac{2}{3N}}}.
$$
Taking room conditions as approximately $T=300$ K and $p=10^5$ Pa, with $V=10^{-6}$ m$^3$, gives
$$
N=\frac{pV}{k_BT}\simeq2.41\times10^{19},\qquad
\boxed{\frac{\sigma_U}{U}\simeq1.66\times10^{-10}}.
$$
The mean <energy> is about $0.15$ J and its <standard deviation> about $2.49\times10^{-11}$ J. The tiny relative fluctuation explains why macroscopic equilibrium <internal energy> behaves as a sharply defined thermodynamic quantity. The estimate uses a <thermal reservoir>, not a particle reservoir; allowing $N$ itself to fluctuate changes the relevant fluctuation problem.
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