Step 1: Model the polarized sphere as two overlapping charged spheres.
A uniformly polarized dielectric sphere can be pictured as a sphere of uniform positive charge density $+\rho$ slightly displaced (by a small vector $\vec{d}$) from an identical sphere of uniform negative charge density $-\rho$, with $\vec{P} = \rho\vec{d}$. Almost all of the charge cancels in the overlap region, and only a thin surface layer of bound charge is left uncancelled, which is exactly the physical picture of bound surface charge on a polarized dielectric.
Step 2: Use the known field inside a single uniform sphere of charge.
Inside a uniformly charged solid sphere of density $\rho$, the field at a displacement $\vec{r}$ from its own center is $\vec{E} = \dfrac{\rho\vec{r}}{3\epsilon_0}$, found from Gauss's law with only the charge inside radius $r$ contributing. At the geometric center of the combined figure, the field from the positive sphere and the field from the negative sphere do not cancel, because they are evaluated relative to two different centers separated by $\vec{d}$.
Step 3: Add the two contributions.
At the common center point, the field from the positive sphere (its own center offset by $-\vec{d}/2$) contributes $\dfrac{\rho}{3\epsilon_0}\big(\vec{d}/2\big)$, and the negative sphere contributes an equal amount in the same net direction. Adding the two gives a total field of magnitude:
\[ E = \dfrac{\rho d}{3\epsilon_0} = \dfrac{P}{3\epsilon_0} \]
directed opposite to $\vec{P}$, confirming the formula from the direct method, now derived from the microscopic bound-charge picture instead of quoted from memory.
Step 4: Put in the numbers.
Converting $P = 26\ \mu\text{C/cm}^2$: since $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$, $P = 26\times10^{-6}/10^{-4} = 0.26\ \text{C/m}^2$. Then:
\[ E = \dfrac{0.26}{3(8.85\times10^{-12})} \approx 9.79\times10^{9}\ \text{N/C} \]
Rounded to one decimal place, $E = 9.8$.
Final Answer:
The field at the center has magnitude $9.8\times10^9\ \text{N/C}$.
\[ \boxed{E = 9.8} \]