The moment of inertia of a plane lamina about an axis is given by the following formula:
$I = \int_R r^2 dm$
where $r$ is the distance from an element of mass $dm$ to the axis, and $R$ is the region of the lamina.
In this case, the region of the lamina is a rectangle with dimensions $2a$ and $b$. The distance from an element of mass $dm$ to the X-axis is $r = x$, where $0 \leq x \leq a$. Therefore, the moment of inertia of the shaded portion of the area about the X-axis is given by:
$I = \int_0^a x^2 b dx$
This integral can be evaluated to give:
$I = \frac{1}{3} ab^2$
Substituting in the values $a = 2$ and $b = 3$ gives:
$I = \frac{1}{3} (2)(3)^2 = 229.34 \text{ cm}^4$
Therefore, the correct answer is $\boxed{A}$.
Option B is incorrect because it is the moment of inertia of the entire rectangle, not just the shaded portion. Option C is incorrect because it is the moment of inertia of a semicircle with radius 2, not the shaded portion. Option D is incorrect because it is the moment of inertia of a quarter circle with radius 2, not the shaded portion.