On the very anniversary of her death, too.

Still, I couldn't expect you to know that. You certainly have a record for not knowing obvious facts and screwing-up numbers.

The following is for my late wife, not for you...

My working:

Imagine a central black hole containing the bulk of the mass of the galaxy.

Sol's orbit distance is 30,000 light years. 1 year is 31 million seconds. Speed of light is 300,000 km/sec

Therefore Sol orbits at 2.79*10^17 km (2.79 *10^20 m) from the centre of mass.

The centripetal acceleration felt by the sun is omega^2 * R or V^2/R. I'll use the omega formulation (angular velocity in radians per second).

Omega is 2*pi/(2.79*10^20*31*10^6) radians per second. = (8.97 * 10^-16)

Centipetal acceleration is Omega^2 * R = 2.24*10^-10 m/sec/sec.

{For interest, given A=GM/R^2, the mass of the galaxy internal to the Sun's orbit is uniquely determined, providing it is a disc, or a sphere, or a point.

M=A/G*R^2 = 2.24*10^-10/6.67*10^-11*(2.79*10^20)^2
= 2.62 * 10^41 Kg

Our Sun masses ~2*10^30 Kg, so the galactic mass (interior to the Sun) is 1.3 * 10^11 Msun. That's 130 billion Suns! (Quick sanity check - one of my "off the cuff" data points is that an average galaxy has 200 billion stars, since the sun is about 2/3 of the way out from the centre, this looks OK)

Doing a quick Excel simulation of that (can't be bothered to work out the integral) gives me a free-fall time of 39.95 Million Years

I leave it as an exercise to the educated and numerate reader to work out alternatives with different mass distributions. I've already explained that a uniform spherical distribution of mass would yield 1/4 of the orbit time (i.e. 56.5 Million Years)

QED

Now bugger off and sort out your maths skills and your interpersonal skills.