1. Electric field due to an infinite line charge

For an infinite straight wire with linear charge density $\lambda$, the electric field at distance $r$ is

$E = \frac{\lambda}{2\pi \varepsilon_0 r}$

Since the electron is revolving around the wire, this electric field provides the required centripetal force.

2. Electrostatic force on the electron

Magnitude of force on electron:

$F_e = eE = e\cdot \frac{\lambda}{2\pi \varepsilon_0 r}$

3. Centripetal force condition

For circular motion,

$\frac{mv^2}{r} = e\cdot \frac{\lambda}{2\pi \varepsilon_0 r}$

Notice that $r$ cancels out:

$mv^2 = \frac{e\lambda}{2\pi \varepsilon_0}$

So,

$v = \sqrt{\frac{e\lambda}{2\pi \varepsilon_0 m}}$

4. Substitute values

Given:

$\lambda = 2\times 10^{-8}\ \mathrm{C\,m^{-1}}$ $e = 1.6\times 10^{-19}\ \mathrm{C}$ $m = 9\times 10^{-31}\ \mathrm{kg}$

Also,

$\frac{1}{4\pi\varepsilon_0}=9\times 10^9$

Hence,

$\frac{1}{2\pi\varepsilon_0}=2\times 9\times 10^9=1.8\times 10^{10}$

Now,

$v^2 = \frac{(1.6\times 10^{-19})(2\times 10^{-8})(1.8\times 10^{10})}{9\times 10^{-31}}$

First simplify numerator:

$1.6\times 2\times 1.8 = 5.76$

and powers:

$10^{-19} \times 10^{-8} \times 10^{10} = 10^{-17}$

So,

$v^2 = \frac{5.76\times 10^{-17}}{9\times 10^{-31}}$

$v^2 = 0.64\times 10^{14} = 6.4\times 10^{13}$

Therefore,

$v = \sqrt{6.4\times 10^{13}}$

$v \approx 8\times 10^6\ \mathrm{m\,s^{-1}}$

5. Final integer value

The question asks for the blank in

$\underline{\hspace{2cm}}\times 10^6\ \mathrm{m\,s^{-1}}$

Hence the required integer is

$\boxed{8}$