The compound shown is treated with excess $\mathrm{LiAlH}{}_4$ in ether, followed by aqueous work-up. Hydride attacks the less hindered carbon of the three-membered ring. Which structure is the final organic product?

The compound is methyl 2-(oxiran-2-yl)acetate. The ester is reduced to $\mathrm{CH}{}_2\mathrm{OH}$ (methanol is lost). Hydride opens the epoxide at the terminal $\mathrm{CH}{}_2$ (SN2 at the less hindered carbon), turning it into a methyl group and leaving the secondary alcohol on the other carbon. The product is butane-1,3-diol. Attack at the more substituted carbon would give butane-1,4-diol.