Explanation:Let edge length of each small cube be \(1 unit\)
Total number of small cubes = \(160 + 56 = 216\)
Volume of big cube = \(216unit^3\)
Side of big cube = \(216^{\frac{1}{3}} = 6 unit\)
Since the exposure of the colored cubes to the outside has to be minimum, we should keep maximum number of colored cubes inside the big cube.
i.e. let us place \(4^3 = 64\) small colored cubes inside the big cube.
Now, exposed cubes = \(216 - 64 = 152\) cubes, out of which \(96\) are colored ones.
T.S.A of big cube = \(6a^2 = 6(6^2) = 216 unit^2\)
Col. A: \(\frac{96}{216}(100) = 44.44%\)
Col. B: \(25.90%\)
Hence, option A
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I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/