= Solution
For a real signal with highest <frequency> $f_{\max}=40\,\mathrm{MHz}$, the <Nyquist–Shannon sampling theorem> gives the nominal Nyquist <sampling rate>
$$
\boxed{f_{s,\mathrm{Nyquist}}=2f_{\max}=80\,\mathrm{MHz}
=80\times10^6\ \text{samples per second}.}
$$
To cover an arbitrary component exactly at the band edge, choose $f_s>80\,\mathrm{MHz}$: a $40\,\mathrm{MHz}$ sine sampled at exactly $80\,\mathrm{MHz}$ can have every sample equal to zero for an unfortunate phase. Practical <aliasing> suppression also calls for a margin above the nominal limiting rate.
With rounding to the nearest digitizer level, <quantization (signal processing)> with step $q=0.1\,\mathrm{mV}$ has the <uniform-quantization error bound>
$$
\boxed{|V_{\mathrm{digitized}}-V_{\mathrm{sample}}|\leq q/2=0.05\,\mathrm{mV}.}
$$
Thus the stated digitizer meets the requested nominal amplitude precision per sample, assuming no clipping or other errors. The bandwidth sets the nominal minimum sampling rate; the level spacing sets the rounding precision. Strictly, the sampling theorem reconstructs exactly from exact samples. A uniform $0.05\,\mathrm{mV}$ error bound on the entire reconstructed continuous waveform is not guaranteed by the level spacing and sampling rate alone; that stronger requirement needs a specified reconstruction and noise model.
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