C

Step 1: Identify the parameters of the biconvex lens.
The radius of curvature (R) = 20 cm, and for a biconvex lens, the focal length (f) is given by the formula:
$$ f = \frac{R}{2} = \frac{20 \text{ cm}}{2} = 10 \text{ cm} $$

Step 2: Use the lens formula to find the image distance (v). The lens formula is given by:
$$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$
where u is the object distance (u = -30 cm).

Substituting the values:
$$ \frac{1}{10} = \frac{1}{v} - \frac{1}{-30} $$
Rearranging gives:
$$ \frac{1}{v} = \frac{1}{10} + \frac{1}{30} $$
Finding a common denominator (30):
$$ \frac{1}{v} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} $$
Inverting gives:
$$ v = \frac{30}{4} = 7.5 \text{ cm} $$

Step 3: Determine the nature of the image.
Since v is positive, the image is real and located on the opposite side of the lens.

Step 4: Calculate the magnification (m) to find the height of the image (h'). The magnification is given by:
$$ m = -\frac{v}{u} = -\frac{7.5}{-30} = 0.25 $$
Thus, the height of the image (h') is:
$$ h' = m \cdot h = 0.25 \cdot 2 \text{ cm} = 0.5 \text{ cm} $$

Step 5: Conclusion.
Therefore, the image is real, inverted, and has a height of 0.5 cm. Thus, the correct answer is Option C.