A bar magnet has a length of 1 cm and a cross-sectional area of 1.0...

Given, length of the magnet, $2l = 1 \text{ cm} = 10^{-2} \text{ m}$
Cross-sectional area, $A = 1.0 \text{ cm}^2 = 10^{-4} \text{ m}^2$
Magnetic field at end-on position, $B = 1.5 \times 10^{-4} \text{ T}$
Distance from the centre, $d = 15 \text{ cm} = 0.15 \text{ m}$

Since $d \gg 2l$, the magnetic field at the end-on (axial) position of a short bar magnet is given by:
$B = \dfrac{\mu_0}{4\pi} \dfrac{2M}{d^3}$

Substituting the given values:
$1.5 \times 10^{-4} = 10^{-7} \times \dfrac{2M}{(0.15)^3}$

$M = \dfrac{1.5 \times 10^{-4} \times 3.375 \times 10^{-3}}{2 \times 10^{-7}}$

$M \approx 2.53 \text{ A m}^2$

The volume of the magnet is:
$V = A \times 2l = 10^{-4} \times 10^{-2} = 10^{-6} \text{ m}^3$

The magnetization $I$ is defined as the magnetic moment per unit volume:
$I = \dfrac{M}{V}$

$I = \dfrac{2.53}{10^{-6}} = 2.53 \times 10^6 \text{ A m}^{-1}$

Rounding to the nearest given option, we get $2.5 \times 10^6 \text{ A m}^{-1}$.