Suppose each iron atom possesses a permanent magnetic moment of 2...

The maximum magnetization $M$ is achieved when all atomic magnetic moments are aligned parallel to each other.

$M = n m$

Substituting the given values:

$n = 8.52 \times 10^{28} \text{ m}^{-3}$

$m = 2 \times 9.27 \times 10^{-24} \text{ A m}^2 = 1.854 \times 10^{-23} \text{ A m}^2$

$M = (8.52 \times 10^{28}) \times (1.854 \times 10^{-23}) \approx 1.58 \times 10^6 \text{ A m}^{-1}$

For a long cylinder, the demagnetizing factor is zero, so the maximum magnetic field on the axis is given by $B = \mu_0 M$.

$B = 4\pi \times 10^{-7} \times 1.58 \times 10^6$

$B = 4 \times 3.1416 \times 0.158 \approx 1.98 \text{ T}$

Rounding to the nearest given option, we get $B \approx 2.0 \text{ T}$.