A

Step 1: Understand the formula for magnifying power (M) of a telescope:
$$ M = \frac{f_o}{f_e} $$
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eyepiece lens.

Step 2: We are given that the magnifying power \( M = 9 \). Therefore:
$$ 9 = \frac{f_o}{f_e} $$
This can be rewritten as:
$$ f_o = 9 f_e $$

Step 3: The total distance between the objective and eyepiece lenses when adjusted for parallel rays is equal to the sum of their focal lengths:
$$ f_o + f_e = 20 \text{ cm} $$

Step 4: Substitute \( f_o \) in the distance equation:
$$ 9 f_e + f_e = 20 \text{ cm} $$
This simplifies to:
$$ 10 f_e = 20 \text{ cm} $$
Therefore:
$$ f_e = 2 \text{ cm} $$

Step 5: Now, substitute \( f_e \) back into the equation to find \( f_o \):
$$ f_o = 9(2) = 18 \text{ cm} $$

Step 6: Therefore, the focal lengths of the objective and eyepiece lenses are \( f_o = 18 \text{ cm} \) and \( f_e = 2 \text{ cm} \) respectively.

Conclusion: The correct option is A: 18 cm, 2 cm.