Upon the insertion of an iron core with a cross-sectional area of 4...

The initial magnetic field inside the solenoid without the core is given by:

$B_0 = \mu_0 H = 2.5 \times 10^{-3} \text{ T}$

The magnetic field after the insertion of the iron core is:

$B = \mu_0 (H + I) = 2.5 \text{ T}$

From the above equations, we can write:

$B = B_0 + \mu_0 I$

Rearranging for the magnetization $I$:

$I = \dfrac{B - B_0}{\mu_0}$

Substituting the given values:

$I = \dfrac{2.5 - 2.5 \times 10^{-3}}{4\pi \times 10^{-7}}$

Since $2.5 \times 10^{-3}$ is very small compared to $2.5$, we can approximate $B - B_0 \approx 2.5 \text{ T}$:

$I \approx \dfrac{2.5}{4 \times 3.14 \times 10^{-7}}$

$I \approx \dfrac{2.5}{12.56 \times 10^{-7}}$

$I \approx 1.99 \times 10^6 \text{ A m}^{-1}$

Rounding to the nearest given option, we get:

$I \approx 2.0 \times 10^6 \text{ A m}^{-1}$