To find the area of the region bounded by the parabola, the tangent at the point (2, 3), and the x-axis, we need to determine the limits of integration and then integrate the area function.

The given equation of the parabola is (y^2)^2 = x. To simplify this equation, we can rewrite it as y^4 = x.

The tangent to the parabola at the point (2, 3) can be found by differentiating the equation of the parabola with respect to x and substituting x = 2 and y = 3.

Differentiating the equation y^4 = x with respect to x, we get:
4y^3 * (dy/dx) = 1

Substituting x = 2 and y = 3, we have:
4(3^3) * (dy/dx) = 1
dy/dx = 1/(4*3^3) = 1/108

So, the equation of the tangent at the point (2, 3) is:
y - 3 = (1/108)(x - 2)

Now we need to find the x-coordinates of the points where the parabola intersects the x-axis. Setting y = 0 in the equation of the parabola, we have:
0^4 = x
x = 0

Therefore, the parabola intersects the x-axis at x = 0.

To find the limits of integration, we need to determine the x-values at which the tangent intersects the x-axis. Setting y = 0 in the equation of the tangent, we have:
0 - 3 = (1/108)(x - 2)
-3 = (1/108)(x - 2)
-324 = x - 2
x = -322

Thus, the tangent intersects the x-axis at x = -322.

We can now set up the integral to find the area of the region bounded by the parabola, the tangent, and the x-axis. Since the parabola is symmetric about the y-axis, we can integrate from x = -322 to x = 0.

The area function is given by:
A = ∫[from -322 to 0] (y^4) dx

But we know that y^4 = x. So the area function becomes:
A = ∫[from -322 to 0] x dx

Integrating x with respect to x, we get:
A = [1/2 * x^2] [from -322 to 0]
A = [1/2 * (0)^2] - [1/2 * (-322)^2]
A = 0 - 1/2 * 103684
A = -51842

Since the area cannot be negative, we take the absolute value:
A = 51842

Therefore, the area of the region bounded by the parabola, the tangent, and the x-axis is 51842 square units.

So the correct answer is option 'C' (9 sq. units).