Answer: $7 \sqrt{2}$.

Solution. On the one hand, the diameter of the sphere cannot be less than the diagonal $14 \sqrt{2}$ of the base (a square with side 14) of the pyramid contained in it, so the radius of the desired sphere is not less than $7 \sqrt{2}$. On the other hand, a sphere with radius $7 \sqrt{2}$ and center at the point of intersection of the diagonals of the base of the pyramid contains its vertex, and with it the entire pyramid, since its height is

$$
h=\sqrt{144-49}=\sqrt{95}<7 \sqrt{2} \quad(\Leftrightarrow 95<98)
$$
3. What is the smallest radius of a sphere from which a regular quadrilateral pyramid with a base edge of 14 and an apothem of 12 can be cut?