Let's denote the edge of the big cube as $a$ and the edges of two not unit as $b$ and $c$.
Then we have an equation:
$$b^3 + c^3 + 2008 = a^3.$$
Because $12^3 = 1728 < 2008$, then $a > 12$, so $a \ge 13$.
Now let's find an upper bound of $a$. It is clear that $a \ge b+c$,
$$b^3 + 3b^2c + 3bc^2 + c^3 \le a^3 = b^3 + c^3 + 2008,$$
which means that $b^2c + bc^2 \le \frac{2008}{3}$ or $b^2c + bc^2 \le 669$. Then $2c^3 \le b^2c + bc^2 \le 669$ so $c^3 \le 334$.

Moreover,
$$2b^2 + 8 \le b^2c + bc^2 \le 669 \Rightarrow 2b^2 \le 661 \Rightarrow b^2 \le 330 \Rightarrow b \le 18 \Rightarrow b^3 \le 5832 \Rightarrow \\ b^3 + c^3 \le 6166 \Rightarrow a^3 = b^3 + c^3 + 2008 \le 6166 + 2008 = 8174 \Rightarrow a \le 20.$$
The only thing that remains is to test on cube of integer numbers $b \ge c \ge 2$, which satisfy an equation $b^3 + c^3 + 2008 = a^3$, where $a$ is from $13$ to $20$. As a result, we have the unique solution: $5^3 + 4^3 + 2008 = 13^3$.