Resolves NO if this has not happened by 2030.
Any reactor in a territory that is claimed by Ukraine counts, regardless of who controls the area.
Please specify what measurement you are using to determine if it's 10^-4 Chernobyls. The ZNPP reactors presumably have very little radioactive iodine in their inventories, having been shut down since September, but there is still enough cesium (and other long lived isotopes) to easily exceed 10^-4 of Chernobyl's cesium release under the wrong conditions.
@ErickBall Total becquerels of radioactive material released into the atmosphere. Doesn't matter what element.
@JonathanRay best numbers I can find put the total Chernobyl release at about 1.4e19 to 1.1e20 Bq. So ZNPP would need to release about 1e16 to resolve yes. That's much less than Three Mile Island (about 4.8e17 Bq), but the total fission product inventory will have fallen by a factor of 10 or more since the reactors were shut down, leaving a total of about 5e18 Bq in each of the 6 reactors (if my math is right). Most of this is non-volatile elements, of which only a tiny fraction would be released in a meltdown. Among volatiles, there should be 3e16 Bq of krypton, 5e17 Bq of cesium, and 2e17 Bq of strontium. If 10% of the krypton and 1% of the cesium and strontium are released (typical of a full meltdown with containment failure) that gets us just barely to 1e16 total. I don't see it happening, not with the cores as cool as they are.
There have been a lot of alarmist articles on this topic e.g https://thehill.com/opinion/national-security/3938345-russia-is-bombing-its-way-toward-nuclear-catastrophe-in-ukraine/amp/ so I create
d this market to find out how exaggerated they are.