 Solution

Use the linear-quadratic radiobiology model, independent fractions with complete inter-fraction repair, and no tumour repopulation. For $N$ fractions of dose $d$, the logarithmic cell kill is $N(\alpha d+\beta d^2)$. Dividing by $\alpha$ defines the biologically effective dose,For the tumour, the original schedule gives $26(2)(1+2/10)=\boxed{62.4{\mathrm{Gy}}_{10}}$. A $3\mathrm{Gy}$ fraction contributes $3(1+3/10)=3.9{\mathrm{Gy}}_{10}$, so precisely $62.4/3.9=\boxed{16}$ such fractions give the same tumour biologically effective dose. The physical total dose changes from $52$ to $48\mathrm{Gy}$.

For the specified normal-tissue $\alpha /\beta =3\mathrm{Gy}$, the original and altered schedules giveThus the normal-tissue model predicts an increase of $9.33{\mathrm{Gy}}_3$, or about $10.8\%$, despite the smaller total physical dose. The smaller normal-tissue $\alpha /\beta$ makes increasing fraction size relatively more costly. These are comparisons within the source's hypothetical radiobiological model and equal-dose geometry, not individualized treatment prescriptions.^[ref]