1. Use the charging relation for a capacitor

For a capacitor, $I = C\,\frac{dV}{dt}$ So, $C = \frac{I}{dV/dt}$

Given: $I = 0.15\ \text{A}, \qquad \frac{dV}{dt} = 7\times 10^8\ \text{V/s}$

Hence, $C = \frac{0.15}{7\times 10^8}$

2. Capacitance of a parallel plate capacitor

For parallel plates, $C = \frac{\varepsilon_0 A}{d}$ where $A = \pi r^2$

Given radius: $r = 10\ \text{cm} = 0.1\ \text{m}$

So area is $A = \pi (0.1)^2 = \pi \times 0.01$

Using $\pi = \frac{22}{7}$, $A = \frac{22}{7}\times 0.01 = \frac{22}{700}\ \text{m}^2$

3. Find distance $d$

From $C = \frac{\varepsilon_0 A}{d}$ we get $d = \frac{\varepsilon_0 A}{C}$

Substitute $C = \frac{I}{dV/dt}$: $d = \varepsilon_0 A\cdot \frac{dV/dt}{I}$

Now put values: $d = 9\times 10^{-12}\times \frac{22}{700}\times \frac{7\times 10^8}{0.15}$

Simplify: $\frac{22}{700}\times 7 = \frac{22}{100} = 0.22$

So, $d = 9\times 10^{-12}\times 0.22\times \frac{10^8}{0.15}$

Also, $9\times 0.22 = 1.98$

Thus, $d = 1.98\times 10^{-12}\times \frac{10^8}{0.15}$

$d = \frac{1.98\times 10^{-4}}{0.15}$

$d = 13.2\times 10^{-4} = 1.32\times 10^{-3}\ \text{m}$

4. Convert to micrometre

Since $1\ \mu\text{m} = 10^{-6}\ \text{m}$

$d = 1.32\times 10^{-3}\ \text{m} = 1.32\times 10^3\ \mu\text{m} = 1320\ \mu\text{m}$

Therefore, the required integer value is $\boxed{1320}$

5. Comparison with stored answer

Stored correct answer = $1320$

This matches the derived result.