1. For a parallel plate capacitor,

$C = \varepsilon_0 \frac{A}{d} = \varepsilon_0 \frac{lb}{d}$

So capacitance is proportional to

$C \propto \frac{lb}{d}$

2. Original capacitor:

$l=3\text{ cm}, \quad b=1\text{ cm}, \quad d=3\,\mu\text{m}$

Thus,

$\frac{lb}{d} = \frac{3\times 1}{3} = 1$

So for the new capacitor to have capacitance increased by a factor of $10$, we need

$\frac{l'b'}{d'} = 10 \times \frac{lb}{d}$

Since the original ratio is $1$, we simply need

$\frac{l'b'}{d'} = 10$

(Here $l,b$ are in cm and $d$ in $\mu$m, so we compare factors relative to the original setup.)

3. Check each case:

Case A

$l=30,\ b=1,\ d=1$

$\frac{lb}{d} = \frac{30\times 1}{1} = 30$

Increase factor $=30$, not $10$.

So A is not correct.

Case B

$l=3,\ b=1,\ d=30$

$\frac{lb}{d} = \frac{3\times 1}{30} = 0.1$

This is one-tenth of original, not ten times.

So B is not correct.

Case C

$l=6,\ b=5,\ d=3$

$\frac{lb}{d} = \frac{6\times 5}{3} = 10$

This is exactly $10$ times the original.

So C is correct.

Case D

$l=1,\ b=1,\ d=10$

$\frac{lb}{d} = \frac{1\times 1}{10} = 0.1$

Not $10$ times.

So D is not correct.

Case E

$l=5,\ b=2,\ d=1$

$\frac{lb}{d} = \frac{5\times 2}{1} = 10$

This is exactly $10$ times the original.

So E is correct.

4. Therefore, the correct cases are:

$\boxed{C \text{ and } E}$

5. Match with given options:

Option D = C and E only.

So the correct option is

$\boxed{\text{D}}$