Usually we can answer questions like this using $ \frac{U}{V} = \frac{B^2}{2 \mu} $ (magnetic energy stored per unit volume), but since iron is nonlinear and ferromagnetic, we need to use its magnetisation curve.

In this case $ \frac{U}{V} = \int_{0}^{B} H \ dB $, so the energy required per unit volume of iron is equal to the area between this curve, the vertical axis, and the line $ B = 1 $. It's quite linear in this region so you could approximate it as a triangle (which is equivalent to just using the original formula above). I estimate it as
$$ \text{energy per unit volume} = \frac{1}{2} \times {1} \times {2000} = 1000 J = 1 \ kJ. $$
Perhaps this diagram looks familiar - we are looking at the dashed part of the curve (initial charging up) in this analysis.