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Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6=4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives $$l^2+l^2+l^2=2^2=4.$$Hence each face has surface area $$l^2=\frac{4}{3}\text{square meters}.$$So the surface area of the inscribed cube is $6\cdot (4/3)=\boxed{8}$ square meters.