10. 3 Let the edge length of the original cube be $a$, and the edge length of the other cube after cutting be $b\left(a, b \in \mathbf{Z}^{+}, a>b\right)$. Then $a^{3}-98=b^{3}$, which means $(a-b)\left(a^{2}+a b+b^{2}\right)=98$.

Thus, $\left\{\begin{array}{l}a-b=1, \\ a^{2}+a b+b^{2}=98 ;\end{array}\right.$ or $\left\{\begin{array}{l}a-b=2, \\ a^{2}+a b+b^{2}=49 ｝ \end{array} \text { or }\left\{\begin{array}{l}a-b=7, \\ a^{2}+a b+b^{2}=14 .\end{array}\right.\end{array}\right.$
Upon inspection, we find that $\quad a=5, b=3$.
10. A cube with an edge length of a certain integer is cut into 99 smaller cubes, 98 of which are unit cubes with an edge length of 1, and the other cube also has an integer edge length. Then its edge length is $\qquad$.