Ground Into Atoms

Let $S$ be the original surface area of the iron cube and $T$ be the total surface area of the smaller cubes. Then each face of the original cube has an area of $\frac{S}{6}$ , and so each of the $10^{2016}$ smaller cube faces would have an area of $\frac{S}{6\sqrt[3]{10^{2016}}^2}$ for a total surface area of $T = 6 \cdot 10^{2016} \cdot \frac{S}{6\sqrt[3]{10^{2016}}^2}$ or $T = 10^{672}S$ .

Since the total surface area increases by $5\underbrace{999999\ldots9}_{671 \text{ nines}}4 = 6 \cdot 10^{672} - 6$ , we have $10^{672}S = 6 \cdot 10^{672} - 6 + S$ , and solving this gives $S = \boxed{6}$ .