= 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>,
$$
\operatorname{BED}=Nd\left(1+\frac d{\alpha/\beta}\right).
$$
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 give
$$
\operatorname{BED}_{\rm old}=52\left(1+\frac23\right)=86.7\,\mathrm{Gy}_3,\qquad \operatorname{BED}_{\rm new}=48\left(1+\frac33\right)=\boxed{96.0\,\mathrm{Gy}_3}.
$$
Thus 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.<Linear-quadratic fractionation study (1989)>[BED formulation and dependence on fraction size and alpha/beta]{ref}
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