What inductance l would be needed to store energy e=3.0kwh (kilowatt-hours) in a coil carrying current i=300a?

The formula for the energy stored in the magnetic field of an inductor is:

                      E  =  (1/2) (inductance) (current)²  .

In the present situation:

Energy = (3 kilo-watt-hour) x (1,000 / kilo) x (joule/watt-sec) x (3,600 sec/hr)

           =  (3 · 1000 · 3,600)  (kilo·watt·hr·joule·sec / kilo·watt·sec·hr)

           =      1.08 x 10⁷ joules .

Now to find the inductance:  

                   E  =  (1/2) (inductance) (current)² 

       (1.08 x 10⁷ joules) = (1/2) (inductance) (300 Amp)²

           (2.16 x10⁷ joules) =  (inductance) (300 Amp)²

             Inductance =  (2.16 x10⁷ joules) / (300 Amp)²

                              =   2.16 x10⁷ / 90,000        Henrys

                           I get        240 Henrys .

This is a big inductance.  Possibly the size of your house.
To get a big inductance, you want to wind the coil
  with a huge number of turns of very fine wire, in
  a small space.
In this case, however, if you plan on running 300A through
  your coil, it'll have to be wound with a very thick conductor ...
  like maybe 1/4-inch solid copper wire, or even copper tubing,
You have competing requirements.
There are cheaper, easier, better ways to store 3 kWh of energy.
In fact, a quick back-of-the-napkin calculation says that
  3 or 4 car batteries will do the job nicely.

That would be kinetic energy

Sublimation- which change of state is shown in model