Imagine drawing the sphere and the cube. Take a cross section, with a plane parallel to two of the cube's faces, passing through the sphere's center. In this cross section, the sphere looks like a circle, and the cube looks like a square (of side length 1) inscribed in that circle. We can now calculate that the sphere has diameter $d:=\sqrt{2}$ and surface area $S:=\pi d^{2}=2 \pi$, and that the sphere protrudes a distance of $x:=\frac{\sqrt{2}-1}{2}$ out from any given face of the cube. It is known that the surface area chopped off from a sphere by any plane is proportional to the perpendicular distance thus chopped off. Thus, each face of the cube chops of a fraction $\frac{x}{d}$ of the sphere's surface. The surface area of the pebble contributed by the sphere is thus $S \cdot\left(1-6 \cdot \frac{x}{d}\right)$, whereas the cube contributes 6 circles of radius $\frac{1}{2}$, with total area $6 \cdot \pi\left(\frac{1}{2}\right)^{2}=\frac{3}{2} \pi$. The pebble's surface area is therefore $$S \cdot\left(1-6 \cdot \frac{x}{d}\right)+\frac{3}{2} \pi=2 \pi \cdot\left(1-6 \cdot \frac{\sqrt{2}-1}{2 \sqrt{2}}\right)+\frac{3}{2} \pi=\frac{6 \sqrt{2}-5}{2} \pi$$