Asserting Run Time From Big O Function
The base of logarithms is irrelevant in big-O notation, since
$$\log_b x = \frac{\log_a x}{\log_a b} = O(\log_a x)\,,$$
so $O(\log_a x)$ and $O(\log_b x)$ denote exactly the same class of functions.

The assertions made by the paper about how long a certain computer would take to solve some problem don't appear to be well founded.  As a minor issue, doesn't "terahertz range computer" mean $10^{12}$ operations per second, not one million per second?  But the major point is that $O(n^2\log n\log\log n)$ is not the same thing as $n^2\log n\log\log n$: it's merely a statement that, there's some constant&nbsp;$c$ such that, for large enough&nbsp;$n$, the number of computation steps is at most $cn^2\log n\log\log n$.  In particular, it doesn't tell us what the value of&nbsp;$c$ is: it could be $10^{-100}$ or $10^{100}$ or anything else. Further, the analysis is never sufficiently precise to equate some number of "computation steps" with an actual number of machine instructions.

You simply cannot turn an asymptotic expression such as this into a running time in seconds without more information.  One possible way to obtain the value of&nbsp;$c$ would be to time the actual execution of the program, but this isn't possible as we don't have any quantum computers that can run Shor's algorithm on $1000$-digit numbers. I don't see anything that would enable the authors of the paper to give precise wall-clock running times for a program on a computer that doesn't exist.

It appears that the authors are using $\ln$ to denote base-$2$ logarithms, since they say that a $k$-digit number equates to $k\ln10$ bits.  A  1000-digit number is about 3300 bits and
$$3300^2\log_2 3300\log_2\log_2 3300\approx 451\times 10^6\,,$$
so that's certainly not where the $84$-second claim comes from.

It seems, in fact, to come from interpreting $\ln$ as meaning the natural logarithm. Based on the incorrect claim that a $1000$-digit number has $1000\log_e 10\approx 2300$ bits, we get
$$2300^2\log_e 2300\log_e\log_e2300\approx 84\times 10^6\,.$$
This calculation is not meaningful in any way.