Conversion from Planck unit to SI Good evening, I'm reading the paper Prehawking radiation by William G. Unruh where it says:

...a time scale of order of $m^{3}$ in Planck units, or $10^{53}$ ages of the current universe for a solar mass black hole"

How do I perform the conversion from Planck units to seconds?
 A: The paper is saying that the time in Planck units is of order $m^3$ where $m$ is the mass of the black hole in Planck units.
The paper uses an example of a Solar mass black hole so $ m = 2 \times 10^{30}$ kg. One Planck mass is $2.18 \times 10^{-8}$ kg, so the mass of the black hole in Planck units is:
$$ m = 9.19 \times 10^{37} $$
and therefore:
$$ m^3 = 7.76 \times 10^{113} $$
So the timescale is of order $7.76 \times 10^{113}$ Planck times. One Planck time is $5.39 \times 10^{-44}$ seconds, so the time in seconds is:
$$\begin{align}
 t &= 4.18 \times 10^{70} \, \mathrm{seconds} \\
   &= 1.33 \times 10^{63} \, \mathrm{years}
\end{align}$$
And since the age of the universe is $1.4 \times 10^{10}$ years the timescale is about $10^{53}$ times greater than the age of the universe, just as it says in the paper.
A: Another way to say what John Rennie said is:
A formula such as $$t=m^3$$ written "in Planck units" means $$\frac{t}{t_P}=\left(\frac{m}{m_P}\right)^3$$ in non-Planck units, where $$t_P=\sqrt{\frac{\hbar G}{c^5}}$$ is the Planck time and $$m_P=\sqrt{\frac{\hbar c}{G}}$$ is the Planck mass.
So in SI units the formula would be
$$t=\frac{t_P}{m_P^3}m^3=\frac{G^2 m^3}{\hbar c^4}.$$
In general, to convert any equation in Planck units to SI units, just divide every mass by the Planck mass, every time by the Planck time, and every length by the Planck length.
