How do you calculate the black hole diameter of M87*? The recent data from the EHT Consortium on the size and mass of the central black hole of M87, named M87*, are telling us that the diameter of the event horizon should be ~1.5 light-days, or stated differently, ~42 ± 3 μas (micro-arc-seconds). They also tell us that the mass is ~ $6.6\times10^9 M_{\odot}$ (billion), and that it is located at a distance of about $16.8 ± 0.8$ Mpc.
How do you calculate the diameter using those fundamentals?
Where $R$ = distance from earth, $M$ = mass of M87* and using SI units please.
 A: The predicted radius of the photon ring is 
$$ r_p = \sqrt{27} \frac{GM_{\rm BH}}{c^2}, $$
for a non-spinning black hole. The result is only slightly different for a fast spinning black hole. By dividing $r_p$ by the distance to the source $D$, we have a relationship between the angular radius and the black hole mass.
This is done in equation 1 of the fifth Event Horizon Telescope paper
$$ \theta_p = \frac{r_p}{D} = 18.8 \left(\frac{M_{\rm BH}}{6.2\times 10^9 M_{\odot}}\right) \left(\frac{D}{16.9\ {\rm Mpc}}\right)^{-1}\ {\rm microarcseconds}\ .$$
If the angular radius is measured to be 21 $\mu$arcsec, then this suggests a mass of 6.9 billion solar masses.
The difference between this and the final result of 6.5 billion solar masses is down to a more sophisticated modelling of the image using a radiative transfer model and a spinning black hole.
A: If you are talking about the diameter of the event horizon, we can use the Schwarzschild radius to get a quick estimate. The Schwarzschild radius is given by 
$$ r_s = \frac{2 G M}{c^2}$$
in SI units and using the given data (we only really need the mass of the black hole which is $M = 6.6 \times 10^9 \times 2 \times 10^{30}  \approx 13.2 \times 10^{39} \text{kg}$ where the mass of the sun is approximately $2 \times 10^{30}$ kg) it is straightforward to calculate the Schwarzschild radius in SI units 
$$ r_s = \frac{2 \times 6.67 \times 10^{-11} \times 13.2 \times 10^{39} }{\left(3 \times 10^8 \right) ^2} \approx 19.6 \times 10^{12} \text{m} $$
A light-day is approximately $25.9 \times 10^{12} \text{m}$ and hence you get that the diameter is approximately 
$$
19.6 \times 2 \div 25.9 \approx 1.51 \ \text{ light-days }
$$
or in SI units if you like 
$$
3.92 \times 10^{13} \text{ m }
$$
I have taken just a few significant figures since the most crucial data (mass of the blackhole) is accurate to at most 2 significant figures. 
