When does Alcohol act as Solvent and as a Nucleophile?
The proposed solution to this roadmap problem has several errors. I am assuming that $\ce{K2Cr2O7}$, methanol and $\ce{LiAlH4}$ are employed in excess. In the oxidation of 1 (A), chromic acid, $\ce{H2Cr2O7}$, is formed from $\ce{K2Cr2O7}$. To obtain the purported hydroxy acid 6 suggested in the solution, exclusive oxidation of the primary alcohol would be required, a pathway that leads to lactone 5 via hemiacetal 4. One could argue that hydrolysis of the lactone occurs to form 6 but, if this were the case, then hydroxy acid 6 would oxidize to levulinic acid 2 in the presence of excess oxidant. Clearly, levulinic acid 2 fits the formula $\ce{C4H8O3}$ presented in the original problem. As the problem is set forth, oxidation of the secondary alcohol is faster than that of the primary alcohol of diol 1.
The Fischer esterification of levulinic acid 2 leads to methyl levulinate 3 ($\ce{C6H10O3}$). Such an esterification is invariably conducted with excess methanol. The reduction of 3 with $\ce{LiAlH4}$ is conducted in a solvent such as ether or tetrahydrofuran (step 1) and then decomposed cautiously with water (step 2) to reform diol 1 (A) by reduction of both the ketone and ester groups of 3. Clearly, this sequence of transformations is an academic exercise and not one to be conducted in a laboratory. Hopefully, the person who provided the solution was not the same person as the one who created the question.