(c) 2 : 3
Surface area of sphere = $4\pi r^2$. Each hemisphere has CSA $= 2\pi r^2$ and a flat circular face $= \pi r^2$, so TSA of one hemisphere $= 3\pi r^2$. Total for two hemispheres $= 6\pi r^2$. Ratio $= 4\pi r^2 : 6\pi r^2 = 2:3$.
When the sphere is cut, each hemisphere gains a flat circular face ($\pi r^2$). So the two hemispheres together have surface area $2(2\pi r^2 + \pi r^2) = 6\pi r^2$, not $4\pi r^2$. The sphere's surface area stays $4\pi r^2$. The key mistake students make is forgetting to add the two flat faces. Always account for newly exposed surfaces when a solid is cut.