A long solenoid has 1000 turns per meter and carries a current of 3...

The percentage increase in the magnetic field is given by $\chi_m \times 100\%$, where $\chi_m$ is the magnetic susceptibility of the core material.

Given that the percentage increase is $4\%$, we have:

$\chi_m = \dfrac{4}{100} = 0.04$

The magnetic intensity $H$ inside the solenoid is determined by the number of turns per unit length $n$ and the current $I$:

$H = nI = 1000 \times 3 = 3000\text{ A/m}$

The magnetization $M$ of the core material is related to the magnetic intensity by:

$M = \chi_m H$

$M = 0.04 \times 3000 = 120\text{ A/m}$

Answer: $120\text{ A/m}$