One. (20 points) Divide a cube with an edge length of a positive integer into 99 smaller cubes, among which, 98 smaller cubes are unit cubes. Find the surface area of the original cube.
Let the side length of the original cube be $x$, and the side lengths of the 99 smaller cubes be 1 and $y(x>y>1)$. Then
$$
\begin{array}{l}
x^{3}-y^{3}=98 \\
\Rightarrow(x-y)\left[(x-y)^{2}+3 x y\right]=98 \\
\Rightarrow(x-y) \mid 98=7^{2} \times 2 \\
\Rightarrow x-y=1,2,7,14,49,98 . \\
\text { By }(x-y)^{3}<x^{3}-y^{3}=98 \\
\Rightarrow x-y=1,2 .
\end{array}
$$

When $x-y=1$, we have $3 x y=97$, which is a contradiction;
When $x-y=2$, we have $x y=15$, solving gives $(x, y)=(5,3)$.
Therefore, the surface area of the original cube is $6 \times 5^{2}=150$.