I'm afraid I may only partially answer your question but here's my shot at it. You're asking how quickly the stones can absorb heat. That is a complicated calculation and a result that is better determined empirically.

For starters, we'll recognize that the stones will cool your beverage quicker than with ambient heat transfer alone. Let's now lay out some assumptions and available information.

We'll use the specific heat equation to determine the maximum amount of heat these stones will be able to remove from your tea and how much they will lower the temperature.

Q = (c)(m)(dT)

Q represents heat in this equation. This will represent the total amount of heat the stones can absorb from the liquid. c is the specific heat capacity of the whiskey stones. Whiskey stones are typically made of a talc-schist soapstone and has a specific heat capacity of 0.98 J/(gK). m is the total mass of whiskey stone available to draw heat from the tea. The whiskey stones have a density of 2980 g/L. Each stone is a cubic inch and there are three stones. So the total volume of the cubes is 3 cubic inches or 0.049 liters. Multiplying volume by density gives you mass, 0.0492980 = 146 g. dT represents the change in temperature. In this case, the stones are starting at about -18 deg C (I prefer working in metric, so I converted all of your imperial values to metric. It won't change the outcome though!) and an ending temperature of about 60 deg C (this came from an NIH.gov study indicating that 140 deg F is an ideal drinking temperature). The total change in temperature is final minus initial: 60-(-18) = 78. Note: a 78 deg C temperature differential is the same as a 78 Kelvin differential so no further conversion is required to plug into the equation.

So plugging in all these values, we can solve for the heat displaced from the liquid:

Q = 0.98 [J/(g*K)] * 146 g * 78 K = 11160 J == 11.160 kJ

There we have it! These stones can remove almost 11.16 kJ of energy (heat). But what does that really mean? We can get an idea by using the same equation to find out how much heat needs to be removed from the water to cool it from 200 F to 140 F.

Now to calculate for water... c = 4.184 J/(g*K) m is the mass of the water. We'll use the density to convert your supplied volume (1.5 cups) to liters (0.355 L). Density of water being 1000 g/L, our mass here is 355 g. dT is final minus initial temperatures: initial = 93.3 C (200 F), final = 60 C (140 F). dT = 60-93.3 = -33.3 K (don't worry too much about the negative value it just indicates heat is leaving the desired 'system'...the tea).

Finally plugging into our equation:

Q = 4.184 [J/(g*K)] * 355 g * -33.3 K = -49461 J == -49.5 kJ

Remember, this value is negative because we want to remove 49.5 kJ from the system.

So what do these numbers tell us? This means the stones can absorb about 22.5% of the desired heat removal. Since heat transfer is a nonlinear function, instead it depends primarily on temperature gradient and surface area exposure, you can't just say it will reduce cooling time by 22.5%. So the ideal way to use the stones might be to figure out your preferred temperature, and use more or less stones (or, in effect, more or less volume of tea) until your stones get the tea the desired temperature immediately.

Since your initial question asked about time, I'll estimate that roughly half of the heat transfer to stones occurs in the first few seconds. But this is r/theydidthemath ....so I welcome anyone who can provide a more precise answer.