= Solution
\b[Build the phenethyl <grignard reagent>, use one portion to make a phenethyl-substituted <ester>, and add two further equivalents to that <ester>.] Write $R=\mathrm{PhCH_2CH_2}$; the target is $\mathrm{R_3COH}$, so all three <carbon> substituents must be phenethyl, not benzyl.
<hydroboration> followed by <oxidation> of <styrene> gives $\mathrm{PhCH_2CH_2OH}$: use $\mathrm{BH_3}$ in a suitable <ether>, followed by alkaline <hydrogen> peroxide. The <Anti-Markovnikov addition> regiochemistry is essential; acid-catalysed <Markovnikov addition> of water would give the wrong <alcohol>. Convert the primary <alcohol> to $\mathrm{PhCH_2CH_2Br}$ with <phosphorus tribromide>, then form $\mathrm{RMgBr}$ with <magnesium> in dry <ether>. The organomagnesium reagent must be prepared and used without a <protic solvent>.
<carboxylate> one portion with <carbon dioxide> and acidify to obtain $\mathrm{RCO_2H}$, then esterify with <ethanol> under acid catalysis to give $\mathrm{RCO_2Et}$. Treat the isolated <ester> with at least two equivalents of another portion of $\mathrm{RMgBr}$ and then perform an aqueous acidic work-up:
$$
\mathrm{RCO_2Et\xrightarrow[\text{then aqueous acid}]{2\,RMgBr}R_3COH}.
$$
The first addition to the <ester> gives a <ketone> after <ethoxide> elimination; the second addition gives the tertiary <alkoxide>. Work-up supplies its OH <proton>. The extra central <carbon> comes from <carbon dioxide>, and each original <styrene> contributes one intact eight-carbon phenethyl unit. The resulting <carbon> count is $1+3\times8=25$, matching the target skeleton.
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