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

Given:
Initial magnetic field, $B_0 = 2.5 \times 10^{-3} \text{ T}$
Final magnetic field, $B = 2.5 \text{ T}$
Cross-sectional area, $A = 4 \text{ cm}^2 = 4 \times 10^{-4} \text{ m}^2$

The magnetic field inside the solenoid without the core is $B_0 = \mu_0 H$.
With the iron core inserted, the magnetic field becomes $B = \mu_0 (H + I)$, where $I$ is the intensity of magnetization of the core.

Therefore, $B = B_0 + \mu_0 I$
$\Rightarrow I = \dfrac{B - B_0}{\mu_0}$

Since $B_0 \ll B$, we can approximate $B - B_0 \approx B = 2.5 \text{ T}$.

The pole strength $m$ developed in the core is given by:
$m = I \times A \approx \dfrac{B}{\mu_0} \times A$

Substituting the given values:
$m = \dfrac{2.5}{4 \pi \times 10^{-7}} \times 4 \times 10^{-4}$

$m = \dfrac{2.5 \times 10^3}{\pi} = \dfrac{2500}{\pi} \approx 795.8 \text{ A m}$

The closest value among the given options is $800 \text{ A m}$.