= Solution
Open the <lactone> formally at its acyl–<oxygen> bond, then replace the resulting secondary <alcohol> by its <ketone> precursor. With $R=\mathrm{CH(Me)CH(Me)_2}$, the useful sequence of retrosynthetic intermediates is
$$
\text{lactone}\ \Longleftarrow\
\mathrm{HO_2CCH(Me)CH_2CH_2CH(OH)R}
\ \Longleftarrow\
\mathrm{EtO_2CCH(Me)CH_2CH_2COR}.
$$
Disconnect the last compound as a <Michael addition> product of the methyl-side <enolate> of $\mathrm{RCOCH_3}$ and <ethyl methacrylate>:
$$
\mathrm{RCOCH_3+CH_2{=}C(Me)CO_2Et}
\longrightarrow
\mathrm{RCOCH_2CH_2CH(Me)CO_2Et}.
$$
The substituted <carbon> of the methacrylate becomes the methyl-bearing <carbon> alpha to the <lactone> <carbonyl>. The <ketone> <carbon> becomes the oxygen-bearing ring <carbon>. This checks the six-membered ring connectivity.
The required branched methyl <ketone> itself comes from <butan-2-one> by ethyl-side <aldol addition> to <acetone>, followed by <dehydration> and selective <hydrogenation>:
$$
\mathrm{MeCOCH_2Me}
\longrightarrow\mathrm{MeCOCH(Me)C(OH)Me_2}
\longrightarrow\mathrm{MeCOC(Me){=}CMe_2}
\longrightarrow\mathrm{MeCOCH(Me)CHMe_2}.
$$
Choose and isolate the aldol regioisomer shown. For the subsequent Michael step, use a bulky base at low temperature to favour its less hindered methyl-side <kinetic enolate>. Reduce the resulting <ketone> selectively with $\mathrm{NaBH_4}$ while retaining the <ester>, hydrolyse the <ester>, then acidify and promote intramolecular <esterification> with water removal. \b[The hydroxy acid closes to the required six-membered <lactone>.] There is no specified stereochemical relationship to enforce, but a stereochemically pure target would require an additional selectivity or resolution step.
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