A

Step 1: For a convex mirror, the focal length (f) is given by $f = \frac{R}{2}$, where R is the radius of curvature.
Therefore, $f = \frac{R}{2}$.
Step 2: For the small circle of radius a placed in front of the convex mirror at a distance equal to the center of curvature (R), the image distance (v1) can be calculated using the mirror formula:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ where u = -R (object distance).
Therefore, $\frac{1}{\frac{R}{2}} = \frac{1}{-R} + \frac{1}{v_1}$.
Simplifying gives us $v_1 = \frac{R}{3}$ (virtual image).
Step 3: The magnification (M) for a spherical mirror is given by $M = -\frac{v}{u}$.
Therefore, $M_1 = -\frac{(R/3)}{-R} = \frac{1}{3}$.
The image radius (r1) due to the first image will be $r_1 = a \cdot M_1 = a \cdot \frac{1}{3}$.
Step 4: Now, this first image acts as an object for the concave mirror at a distance of 2R - (R/3) = (6R/3) - (R/3) = (5R/3). Apply the same mirror formula for the concave mirror:
Using $u = -\frac{5R}{3}$, we calculate $v_2$ using the formula. After simplification, it gives $v_2 = \frac{5R}{6}$. Thus, the second image radius (r2) can be calculated as
$r_2 = a \cdot M_2 = a \cdot \frac{5}{6}$.
Step 5: The second image again acts as an object for the convex mirror, and following the same process, we find the distance becomes (8R/6) and thus we find the third image so that
$r_3 = a \cdot M_3 = a \cdot \frac{27}{32}$.
The final resultant image sizes are $r_1, r_2, r_3$.
Therefore, the ratio of image radii corresponding to the first three images will be \{\frac{1}{3}, \frac{5}{6}, \frac{27}{32}\}.
Therefore, the answer is A.