= Solution
Write the proton-labelled structure as $\mathrm{CH_3^{a}C(=O)CH^{b}(Cl)CH_3^{c}}$. The assignments are
$$
\begin{array}{c|c|c|c}
\text{site}&\delta_H/\mathrm{ppm}&\text{integral}&\text{multiplicity}\\\hline
a&2.334&3&\text{singlet}\\
b&4.3365&1&\text{quartet}\\
c&1.6065&3&\text{doublet}
\end{array}
$$
The <NMR quartet> is centred at $(4.371+4.302)/2=4.3365$ ppm; each adjacent line differs by $0.023$ ppm. The <NMR doublet> has the same spacing, $1.618-1.595=0.023$ ppm. At 300 MHz,
$$
\boxed{{}^3J_{bc}=300(0.023)=6.9\,\mathrm{Hz}.}
$$
Three equivalent methyl <protons> split $b$ into a $1:3:3:1$ <NMR quartet>; the single methine <proton> splits $c$ into a $1:1$ <NMR doublet>. No resolved <long-range scalar coupling> splits the acetyl methyl, giving its <NMR singlet>. The integrals total seven <protons>.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-chloroketone-assignment.png]
{title=Independent original carbon and proton assignments for A, with the measured methine–methyl coupling}
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