Extra Credit: Ch 14 #46:
We use Newton's version of Kepler's third law:

P² = a³ * 4 pi² / G (M1 + M2)

Let M1 be the mass of the black hole. M2 is the mass of the star.
M1 is much, much, much greater than M2, so we can safely ignore M2. In other words, you can let M2 = 0.

For a circular orbit, the circumference is (2 * pi * a).
The time it takes to make an orbit is the distance / speed, which is the circumference divided by the velocity.
So P = (2*pi*a) / v .
And now we can go back to Kepler's 3rd law and solve for M1.

You can check your algebra, since the answer is given in the text book:
M1 = a * v² / G

Now we just need to plug in the numbers. But first, we have to make sure the units make sense.
Convert the given quantities to the standard metric system:
v = 1000 km/s = 10⁶ m/s
a = 20 light days = 5.184x10¹⁴ m


Plugging in (and after cancelling the units) we have:
M1 (in kg) = 5.184x10¹⁴ (10⁶)² / 6.67x10^-11
M1 (in kg) = 7.77x10³⁶

And thus:
M1 = 3.91x10⁶ Solar Masses
You have determined the mass of the supermassive black hole Sag A* !

The solutions presented on this webpage are based on the work of Mr. R. Pandya.