1. Initial capacitance

For a parallel plate capacitor, $C = \frac{\varepsilon_0 A}{d}$ when air is between the plates.

Given: $C_1 = 15\ \text{pF}$

2. Changes made to the capacitor

• Plate separation is doubled: $d' = 2d$
• Dielectric of constant $K = 3.5$ is inserted.

Now the new capacitance is $C_2 = \frac{K\varepsilon_0 A}{d'} = \frac{K\varepsilon_0 A}{2d}$

Using $C_1 = \frac{\varepsilon_0 A}{d}$, $C_2 = \frac{K}{2} C_1$

3. Substitute values

$C_2 = \frac{3.5}{2} \times 15$

$C_2 = 1.75 \times 15 = 26.25\ \text{pF}$

4. Match with the given form

Given that the new capacitance is $\frac{x}{4}\ \text{pF}$

So, $\frac{x}{4} = 26.25$

$x = 26.25 \times 4 = 105$

5. Final answer

$\boxed{105}$