= Solution
For the $110\,\mathrm{Hz}$ fundamental alone, the <Nyquist–Shannon sampling theorem> places the limiting <sampling rate> at twice that <frequency>:
$$
\boxed{f_s\gtrsim2(110\,\mathrm{Hz})=220\,\mathrm{samples\,s^{-1}}.}
$$
Use a rate strictly above this boundary for a sinusoid of unknown <phase>: exactly two samples per cycle can, for a particular <phase>, all fall at zero. If higher <Fourier harmonics> are retained, the <sampling rate> must instead exceed twice the highest retained <frequency>, or those components must be removed to prevent <aliasing>.
For a straightforward <discrete Fourier transform> estimate, <Fourier frequency resolution> obeys $\Delta f\sim1/T_{\mathrm{obs}}$. Resolving $0.07\,\mathrm{Hz}$ therefore suggests
$$
\boxed{T_{\mathrm{obs}}\gtrsim\frac1{0.07\,\mathrm{Hz}}=14.3\,\mathrm s.}
$$
At the limiting <sampling rate> this is about $f_sT_{\mathrm{obs}}\simeq3143$ samples. These are the usual sampling and spectral-resolution estimates. The $14.3\,\mathrm s$ value is not a universal lower bound on single-tone <frequency estimation>: fitting a sinusoidal model with sufficiently low noise can estimate its <frequency> more precisely than the spacing of <discrete Fourier transform> bins. The question gives no noise level or estimator, so a sharper universal minimum time cannot be deduced.
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