13. Given $a, b, c \in \mathbf{Z}, a, b, c > 1$, satisfying: $\sqrt[a]{N \sqrt[b]{N \sqrt[c]{N}}}=\sqrt[36]{N^{25}}(N>1)$. Then the value of $b$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
13. B.

From the problem, we know
$$
\begin{array}{l}
N^{\frac{1}{a}+\frac{1}{a b}+\frac{1}{a x}}=\sqrt[a]{N \sqrt[b]{N \sqrt[c]{N}}}=\sqrt[36]{N^{25}}=N^{\frac{25}{3^{3}}} \\
\Rightarrow \frac{b c+c+1}{a b c}=\frac{25}{36} .
\end{array}
$$

If $a \geqslant 3$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{c+1}{3 b c}=\frac{b c+c+1}{3 b c} \\
>\frac{b c+c+1}{a b c}=\frac{25}{36}>\frac{21}{36}=\frac{7}{12} \\
\Rightarrow \frac{c+1}{3 b c}>\frac{1}{4} \\
\Rightarrow 4 c+4>3 b c \geqslant 6 c \\
\Rightarrow c<4 \\
\Rightarrow c=1, 2, 3.
\end{array}
$$
If $c=1$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{2}{3 b}=\frac{b+2}{3 b} \\
>\frac{25}{36} \\
\Rightarrow \frac{2}{3 b}>\frac{1}{12} \\
\Rightarrow 8>3 b \\
\Rightarrow b<3 \\
\Rightarrow b=1, 2.
\end{array}
$$
If $b=1$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{2}{3}=\frac{1}{3}+\frac{2}{3}=\frac{3}{3}=1 \\
\Rightarrow 1>\frac{25}{36} \text{ (contradiction)}.
\end{array}
$$
If $b=2$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{2}{6}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3} \\
\Rightarrow \frac{2}{3}>\frac{25}{36} \text{ (contradiction)}.
\end{array}
$$
If $c=2$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{3}{6 b}=\frac{2 b+3}{6 b} \\
>\frac{25}{36} \\
\Rightarrow \frac{3}{6 b}>\frac{1}{12} \\
\Rightarrow 6>3 b \\
\Rightarrow b<2 \\
\Rightarrow b=1.
\end{array}
$$
If $b=1$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{3}{6}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6} \\
\Rightarrow \frac{5}{6}>\frac{25}{36} \text{ (contradiction)}.
\end{array}
$$
If $c=3$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{4}{9 b}=\frac{3 b+4}{9 b} \\
>\frac{25}{36} \\
\Rightarrow \frac{4}{9 b}>\frac{1}{12} \\
\Rightarrow 16>9 b \\
\Rightarrow b<\frac{16}{9} \\
\Rightarrow b=1.
\end{array}
$$
If $b=1$, then
$$
\begin{array}{l}
\frac{1}{3}+\frac{4}{9}=\frac{3+4}{9}=\frac{7}{9} \\
\Rightarrow \frac{7}{9}>\frac{25}{36} \text{ (contradiction)}.
\end{array}
$$
Thus, $a=2$.
So, $7 b c=18 c+18$
$$
\Rightarrow c=\frac{18}{7 b-18} \Rightarrow b=3 .
$$