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(B) Initial surface area of the cube with edge length $L_1 = 1 \ cm$ is $A_1 = 6 	imes (L_1)^2 = 6 \ cm^2$. The edge length of the smaller cube is $L

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$10^7$

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(B) Initial surface area of the cube with edge length $L_1 = 1 \ cm$ is $A_1 = 6 	imes (L_1)^2 = 6 \ cm^2$. The edge length of the smaller cube is $L_2 = 1 \ nm = 10^{-7} \ cm$. The number of smaller cubes $n$ is given by the ratio of volumes: $n = \frac{(L_1)^3}{(L_2)^3} = \frac{1^3}{(10^{-7})^3} = 10^{21}$. The total surface area of all $n$ smaller cubes is $A_2 = n 	imes 6 	imes (L_2)^2 = 10^{21} 	imes 6 	imes (10^{-7})^2 = 10^{21} 	imes 6 	imes 10^{-14} = 6 	imes 10^7 \ cm^2$. The ratio of the total surface area of the smaller cubes to the initial cube is $\frac{A_2}{A_1} = \frac{6 	imes 10^7}{6} = 10^7$.

$A$ cube of edge length $1 \ cm$ is divided into smaller cubes of uniform size of length $1 \ nm$. Assuming that no voids are present,the ratio of total surface area of all the cubes of $1 \ nm$ edge length to the surface area of the initial cube is

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