A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are:

For two thin lenses of focal lengths $$f_1$$ and $$f_2$$ kept with their principal axes coincident and separated by a distance $$d$$, the combined (or equivalent) focal length $$F$$ is given by the lens-combination formula $$ \frac{1}{F} \;=\; \frac{1}{f_1} \;+\; \frac{1}{f_2} \;-\; \frac{d}{f_1\,f_2}. $$

Here the numerical data are $$F = 10\ \text{cm}, \qquad d = 2\ \text{cm}.$$ Substituting these values we get $$ \frac{1}{10} \;=\; \frac{1}{f_1} \;+\; \frac{1}{f_2} \;-\; \frac{2}{f_1\,f_2}. $$

To remove the denominators we multiply every term by $$f_1\,f_2$$ and obtain $$ \frac{f_1\,f_2}{10} \;=\; f_2 \;+\; f_1 \;-\; 2. $$ Rearranging makes the relation clearer: $$ f_1\,f_2 \;=\; 10\bigl(f_1 + f_2 - 2\bigr). $$

Now we test each given pair of focal lengths against this necessary condition.

Option A: $$f_1 = 18\ \text{cm},\; f_2 = 20\ \text{cm}$$ Left side: $$f_1\,f_2 = 18 \times 20 = 360.$$ Right side: $$10\bigl(f_1 + f_2 - 2\bigr) = 10\bigl(18 + 20 - 2\bigr) = 10 \times 36 = 360.$$ Both sides are equal, so Option A satisfies the combination formula exactly.

Option B: $$f_1 = 10\ \text{cm},\; f_2 = 12\ \text{cm}$$ Left: $$10 \times 12 = 120$$ Right: $$10\bigl(10 + 12 - 2\bigr) = 10 \times 20 = 200,$$ which is not equal to 120, so Option B is rejected.

Option C: $$f_1 = 12\ \text{cm},\; f_2 = 14\ \text{cm}$$ Left: $$12 \times 14 = 168$$ Right: $$10\bigl(12 + 14 - 2\bigr) = 10 \times 24 = 240,$$ not equal, hence Option C is rejected.

Option D: $$f_1 = 16\ \text{cm},\; f_2 = 18\ \text{cm}$$ Left: $$16 \times 18 = 288$$ Right: $$10\bigl(16 + 18 - 2\bigr) = 10 \times 32 = 320,$$ again unequal, so Option D is rejected.

Only Option A satisfies the required mathematical condition, meaning the focal lengths must be 18 cm and 20 cm.

Hence, the correct answer is Option A.