Can I get a 200mH inductor off the shelf that I can connect across AC mains?
If you assume that the inductor core will be silicon steel laminate you can find two things from the internet: -


The relative magnetic permeability is about 4000 and
The maximum flux density is about 1 tesla for decent inductor operation.




From this you can calculate the maximum H-field applied to the core that you can tolerate before exceeding the maximum flux density. It's simply a case of converting relative to absolute permeability and dividing this into 1 tesla. 

I get a peak H field of 198.9 ampere turns per metre. It pretty much "aligns" with the above graph for silicon steel.

The "per metre" part is the overall mean length of the core: -



Knowing the inductance, frequency and applied voltage you can now calculate the peak current. If I assume 220 volts RMS and 50 Hz, the peak current is 4.952 amps.

Given that you have previously calculated the peak H-field you can divide this by peak current to get "turns per metre" (about 40.2 turns for a 1 metre core length). 

Now the inductance formula: -




A = cross sectional area of core and 
l is the mean length of the core. 
N is number of turns.


If you substitute N/l with H and then divide by the peak current you get: -

\$L = \dfrac{\mu N HA}{I}\$

And, if you plug the numbers in for the known values you get \$AN\$ = 1. This means that for 40 turns, A = 0.025 sq metres and of course this is a very fat core of cross section 158 mm x 158 mm. Additionally, the core length needs to be 1 metre long.

If you were expecting some inch cube or smaller inductor then I hope you can see why it needs to be many times this size. The only possibility of reducing the core size is to gap the core. Gapping the core means the effective permeability drops and you can tolerate a bigger H field. To restore the inductance you have to increase the number of turns of course.

If the effective permeability dropped by 4 (due to gapping), the number of turns would have to double to restore the inductance (because inductance is proportional to turns squared). Now we need 80.4 turns and our peak H-field can now be as high as 795.8 At/m (to produce 1 telsa in the core).

Based on that H-field, the turns per metre is 160.8 or, putting it another way, the length of the core can now be 0.5 metres - this gives us 80.4 turns.

Plugging these into the formula for inductance and re-arranging, the cross sectional area halves from the previous 0.025 sq m to 0.125 sq m. That's 112 mm x 112 mm and a tad more reasonable.

You could keep going with this - making the gap a bit bigger, reducing the effective permeability, increasing the number of turns to restore the inductance but you will get to a point where you have too many turns and your basic \$I^2R\$ losses become too great.

So, apologies for thinking you could get away with an ungapped core (you can but it will be very big) and good luck.