smc

$\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ can be simplified to $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.$ The equation is then $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$ which implies that $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$ $a$ has to be $2$ since $\frac{25}{36}>\frac{7}{12}$. $\frac{7}{12}$ is the result when $a,b,$ and $c$ are $3,2,$ and $2$ $b$ being $3$ will make the fraction $\frac{2}{3}$ which is close to $\frac{25}{36}$. Finally, with $c$ being $6$, the fraction becomes $\frac{25}{36}$. In this case $a,b,$ and $c$ work, which means that $b$ must equal $\boxed{3}$~lopkiloinm