Calculate number of small cubes making up large cube given number in outermost layer

Let the big cube be of dimension $(x+2)$ (made up of $(x+2)^3$ smaller cubes). Then $(x+2)^3-x^3=100,614,152$. This reduces to a quadratic equation which you can solve.

Let $x$ be the number of small cubes along each edge of the large cube. Then each face of the large cube contains $x^2$ small cubes. $6x^2$ isn't the total number of cubes around the outside, though, because we're double-counting the cubes along each of the edges, so we need to subtract off $12x$. Then we aren't counting the cubes at the corners (we counted them three times--once in each face--and subtracted them three times), so we need to add back on 8. So we have $$6x^2 -12x + 8 = 100,614,152.$$ Now this is just a simple quadratic. Combine the terms on one side and use the quadratic formula (or Wolfram Alpha) to find that $x = 4096$.

If one big cube is divided into $n$ times smaller cubes, then there are $n^3$ cubes. Taking avay the outer layer, we are left with $(n-2)^3$ cubes. The difference is $$ n^3-(n-2)^3= n^3-(n^3-6n^2+12n-8)=6n^2-12n+8$$ and this shall equal $N=100614152$. Therefore, a good approximation for $n$ is given by $$n=\sqrt {\frac N 6}\approx 4095.0004$$ However, $n=4095$ leads to $6n^2-12n+8=100565018$, not quite your expected result. But with $n=4096$, the result is correct: $6n^2-12n+8=100614152$.

Note that trying to compute the solution of $$6n^2-12n+(8- 100614152)$$ would not have been easy due to rounding errors.