1. Given data

• Radius of curvature of each mirror: $R = 12\text{ cm}$
• Therefore focal length magnitude: $f = \frac{R}{2} = 6\text{ cm}$
• Object distance from each mirror: $18\text{ cm}$

Using Cartesian sign convention:

• For a convex mirror: $f = +6\text{ cm}$
• For a concave mirror: $f = -6\text{ cm}$
• For both mirrors, object is in front, so $u = -18\text{ cm}$

2. Image formed by convex mirror

Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ Substitute $f=+6$, $u=-18$: $\frac{1}{6} = \frac{1}{v} - \frac{1}{18}$ $\frac{1}{v} = \frac{1}{6} + \frac{1}{18} = \frac{3+1}{18} = \frac{4}{18} = \frac{2}{9}$ $v = \frac{9}{2} = 4.5\text{ cm}$

Magnification for mirror: $m = -\frac{v}{u}$ So, $m_{\text{convex}} = -\frac{4.5}{-18} = \frac{1}{4}$ Thus image size by convex mirror is $\frac{1}{4}$ times the object size.

3. Image formed by concave mirror

Now $f=-6$, $u=-18$: $\frac{1}{-6} = \frac{1}{v} + \frac{1}{-18}$ $-\frac{1}{6} = \frac{1}{v} - \frac{1}{18}$ $\frac{1}{v} = -\frac{1}{6} + \frac{1}{18} = \frac{-3+1}{18} = -\frac{2}{18} = -\frac{1}{9}$ $v = -9\text{ cm}$

Then, $m_{\text{concave}} = -\frac{v}{u} = -\frac{-9}{-18} = -\frac{1}{2}$ The image is inverted, so its size is given by magnitude: $|m_{\text{concave}}| = \frac{1}{2}$

4. Required ratio of image sizes

Ratio of sizes formed by convex mirror to concave mirror: $\frac{\text{size by convex}}{\text{size by concave}} = \frac{1/4}{1/2} = \frac{1}{2}$

5. Final answer

The required ratio is: $\boxed{\frac{1}{2}}$ So the correct option is B.