The area of a circle (in square inches) is numerically larger than its circumference (in inches). What is the smallest possible integral area of the circle, in square inches?

[i]Proposed by James Lin[/i]
1. Let \( r \) be the radius of the circle. The area \( A \) of the circle is given by:
   \[
   A = \pi r^2
   \]
   The circumference \( C \) of the circle is given by:
   \[
   C = 2\pi r
   \]

2. According to the problem, the area of the circle is numerically larger than its circumference:
   \[
   \pi r^2 > 2\pi r
   \]

3. We can divide both sides of the inequality by \(\pi\) (since \(\pi > 0\)):
   \[
   r^2 > 2r
   \]

4. To solve this inequality, we can rearrange it:
   \[
   r^2 - 2r > 0
   \]

5. Factor the quadratic expression:
   \[
   r(r - 2) > 0
   \]

6. This inequality holds when \( r > 2 \) or \( r < 0 \). Since \( r \) represents the radius of a circle, it must be positive. Therefore, we have:
   \[
   r > 2
   \]

7. The smallest integer value of \( r \) that satisfies \( r > 2 \) is \( r = 3 \).

8. Calculate the area of the circle when \( r = 3 \):
   \[
   A = \pi r^2 = \pi (3^2) = 9\pi
   \]

9. To find the smallest possible integral area, we approximate \( 9\pi \):
   \[
   9\pi \approx 9 \times 3.14159 \approx 28.27431
   \]

10. The smallest integer greater than \( 28.27431 \) is \( 29 \).

The final answer is \(\boxed{29}\).