The capacitance of a parallel plate capacitor is given by

$$ C = \frac{\epsilon_0 A}{d}, $$

where:

$$\epsilon_0$$ is the permittivity of free space,

$$A$$ is the area of a plate, and

$$d$$ is the separation between the plates.

A change in the capacitor’s dimensions will change its capacitance by a factor of

$$ \frac{C_{\text{new}}}{C_{\text{initial}}} = \frac{A_{\text{new}}}{A_{\text{initial}}} \cdot \frac{d_{\text{initial}}}{d_{\text{new}}}. $$

The initial dimensions are:

Length: $$l = 3\,\text{cm} = 0.03\,\text{m},$$

Breadth: $$b = 1\,\text{cm} = 0.01\,\text{m},$$

Hence, the original area is

$$ A_{\text{initial}} = 0.03 \times 0.01 = 3 \times 10^{-4}\,\text{m}^2, $$

Plate separation: $$d_{\text{initial}} = 3\,\mu\text{m} = 3 \times 10^{-6}\,\text{m}.$$

For each modification, we compare the new factor:

Case C:

New dimensions:

$$l = 6\,\text{cm} = 0.06\,\text{m}, \quad b = 5\,\text{cm} = 0.05\,\text{m}.$$

New area:

$$ A_{\text{new}} = 0.06 \times 0.05 = 3 \times 10^{-3}\,\text{m}^2. $$

Ratio of areas:

$$ \frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{3 \times 10^{-3}}{3 \times 10^{-4}} = 10. $$

The plate separation is unchanged:

$$ \frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{3 \times 10^{-6}} = 1. $$

Overall factor:

$$ 10 \times 1 = 10. $$

Case E:

New dimensions:

$$l = 5\,\text{cm} = 0.05\,\text{m}, \quad b = 2\,\text{cm} = 0.02\,\text{m}.$$

New area:

$$ A_{\text{new}} = 0.05 \times 0.02 = 1 \times 10^{-3}\,\text{m}^2. $$

Ratio of areas:

$$ \frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{1 \times 10^{-3}}{3 \times 10^{-4}} \approx 3.33. $$

New plate separation:

$$d_{\text{new}} = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m},$$

so the ratio of separations is

$$ \frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{1 \times 10^{-6}} = 3. $$

Overall factor:

$$ 3.33 \times 3 \approx 10. $$

Both Case C and Case E yield a capacitance that is 10 times the original value.

Thus, the correct modifications are those in Case C and Case E.