The magnetic induction produced at the centre of the air-core...

The magnetic induction at the centre of an air-core solenoid is given by

$B_0 = \mu_0 n I$

When the air-core is replaced by an iron-core of susceptibility $\chi$, the magnetic induction becomes

$B_1 = \mu n I$

The permeability of the core $\mu$ is related to the relative permeability $\mu_r$ and the permeability of free space $\mu_0$ by

$\mu = \mu_0 \mu_r$

The relative permeability is related to the magnetic susceptibility $\chi$ by

$\mu_r = 1 + \chi$

Substituting this into the expression for $B_1$, we get

$B_1 = \mu_0 (1 + \chi) n I$

The ratio of the magnetic inductions is

$\dfrac{B_1}{B_0} = \dfrac{\mu_0 (1 + \chi) n I}{\mu_0 n I} = 1 + \chi$

Answer: $1 + \chi$