Let $A=21982145917308330487013369=B^{13}$. Because $A$ has 26 digits, therefore, we must have $$10^{25} < A < 10^{26} \implies B<100$$ Next, we can find the lower boundary of $B$ by noting $$80^{13} < 10^{25} \implies 80 < B < 100$$ Clearly, the last digit of $B$ must be $9$. Hence, $B$ can be only $89$ or $99$. The final step is to check whether $A$ is a multiple of $9$. Clearly, it is not. Hence, $B$ can only be $\boxed{89}$.