The area of the region bounded by the parabola $y^2=27 x$ and the line $x=1$ is ______ sq.units.

The parabola $y^2 = 27x$ opens to the right and is symmetric about the x-axis. The region bounded by this parabola and the line $x = 1$ extends from the vertex at $x = 0$ to the boundary line.

From the parabola equation, $y = \pm 3\sqrt{3}\sqrt{x}$. By symmetry, we compute the area as twice the area above the x-axis:

$A = 2\int_0^1 3\sqrt{3}\sqrt{x}dx = 6\sqrt{3}\int_0^1 x^{1/2}dx$

$A = 6\sqrt{3}\left[\frac{2}{3}x^{3/2}\right]_0^1 = 6\sqrt{3} \cdot \frac{2}{3} = 4\sqrt{3}$

The area is $4\sqrt{3}$ square units, corresponding to option C.