TL;DR

Measurement of mass geometrically?!

I am uncertain exactly how you're making this measurement, but as long as you are not trying to be practical, the answer is probably yes.
For example, if you carefully measured the circumference of a sea-level great circle path around the Earth's equator and then came up with some way of deducing the radius, the ratio would not be exactly 2pi and if you took into consideration the method you used for measuring the radius and the effects of special relativity, you could deduce the Mass of the Earth.

With regards to Sagittarius alpha, do you have a way of measuring a radius and circumference in the vicinity of alpha (aka Rukbat) without visiting it?

If you build two concentric rings of different sizes around a spherical non-rotating mass ##M## and measure their circumferences ##C_1## and ##C_2## then their radial coordinates are defined to be ##R_1=C_1/2\pi## and ##R_2=C_2/2\pi##. If you then build a ladder from one ring to the other, the length of the ladder will be $$\begin{eqnarray*}
&&\int_{R_1}^{R_2}\frac{dr}{\sqrt{1-\frac{R_S}{r}}}\\
&=&\frac{R_S}2\ln\left(
\frac{
\left(1-\sqrt{1-\frac{R_S}{R_1}}\right)
\left(1+\sqrt{1-\frac{R_S}{R_2}}\right)
}{
\left(1+\sqrt{1-\frac{R_S}{R_1}}\right)
\left(1-\sqrt{1-\frac{R_S}{R_2}}\right)
}\right)\\
&&+\sqrt{R_2(R_2-R_S)}-\sqrt{R_1(R_1-R_S)}
\end{eqnarray*}
$$where ##R_S=\frac{2GM}{c^2}## is the Schwarzschild radius of the mass. Note that when ##R_S## is negligible compared to ##R_1## and ##R_2## only the last two terms are significant and they reduce to ##R_2-R_1##. Also note that for Earth, ##R_S\approx 1.5\mathrm{cm}## and the minimum value for ##R_1## is about 6400km, some eight orders of magnitude larger. So while it is possible to deduce ##R_S## and hence ##M## from such calculations, the measurements required to get an answer distinguishable from zero would need to be impossibly precise.

This is, by the way, the physical meaning of the dent in those daft "gravity is like a dent in a rubber sheet" pictures. If you were to build multiple coplanar rings around Earth connected by radial pillars and get Airfix to build a scale model suitable for a child's bedroom (1:10⁸ would be about right), then the radial pillars would be very slightly too long to fit between the rings (if measurements and manufacturing were precise enough anyway) due to the difference between the Schwarzschild geometry behind the original and the (near) Euclidean geometry behind the model. The slightly-too-long pillars would force the model's rings out of the plane, and the shape they would form is the shape of the dent. Which is much shallower for Earth than is typically shown (1mm rise for every few hundred kilometers of run), and has absolutely nothing to do with the everyday gravitational "force".

You could use the maths above for a back-of-the-envelope calculation for the same scenario around SagA*. But it's a Kerr black hole, so formally you'd have to replace the integrand with the square root of the ##g_{rr}## term in the Kerr metric, which may or may not be analytically integrable, and you'd probably have to stay clear of the ergosphere. And I'm not sure how precisely we know its spin parameter.

This is essentially the physical realization of the Riemannian curvature integral. In a locally curved space, the circumference of a circle is indeed not exactly 2\pi r, and the deviation can be quantified as an integral of the curvature tensor over the enclosed area.

Interestingly, this exact geometric deviation (the 'curvature residual') can be parameterized as a dimensionless correction term. In some recent topological frameworks, this residual is denoted as \Delta\pi, representing the local distortion of the spacetime metric from a perfect Euclidean geometry.