= Solution
Take room temperature as $T=300\,\mathrm K$. With the supplied <gyromagnetic ratio>, <reduced Planck constant> and <Boltzmann constant>,
$$
\Delta E=(1.054\times10^{-34})(2.675\times10^8)(1.5)=4.23\times10^{-26}\,\mathrm J,
$$
so
$$
\boxed{\frac{\Delta E}{k_{\rm B}T}=\frac{4.23\times10^{-26}}{(1.380\times10^{-23})(300)}=1.02\times10^{-5}.}
$$
At $293\,\mathrm K$ the ratio is $1.05\times10^{-5}$; the room-temperature convention makes only a small difference. The corresponding fractional <nuclear spin polarization> has magnitude about five parts per million. The source's negative sign convention assigns the label up to the higher-energy state; it should not be confused with the lower-energy magnetic moment aligned with a positive proton <magnetic field>.
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