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.