1. Given data

• Intensity of radiation:
  $I = 6.4\times 10^{-5}\ \text{W/cm}^2$
• Area of metal surface:
  $A = 1\ \text{cm}^2$
• Wavelength:
  $\lambda = 310\ \text{nm}$
• Work function:
  $\phi = 2\ \text{eV}$
• One in $10^3$ photons ejects one electron.

We need total number of electrons emitted in $1\ \text{s}$.

2. Power falling on the surface

Since area is $1\ \text{cm}^2$, $P = IA = 6.4\times 10^{-5}\ \text{W}$

So, energy incident in $1\ \text{s}$ is $E_{\text{total}} = P\times 1 = 6.4\times 10^{-5}\ \text{J}$

3. Energy of one photon

Using $E = \frac{hc}{\lambda}$ Given $hc = 1240\ \text{eV nm}$, $E_{\gamma} = \frac{1240}{310}\ \text{eV} = 4\ \text{eV}$

Convert into joules: $E_{\gamma} = 4\times 1.6\times 10^{-19} = 6.4\times 10^{-19}\ \text{J}$

4. Check whether photoemission occurs

Photon energy = $4\ \text{eV}$ and work function = $2\ \text{eV}$.

Since $4 > 2,$ photoelectrons are emitted.

5. Number of photons incident per second

$N_{\gamma} = \frac{E_{\text{total}}}{E_{\gamma}} = \frac{6.4\times 10^{-5}}{6.4\times 10^{-19}} = 10^{14}$

So, incident photons per second = $10^{14}$

6. Number of electrons emitted per second

Given only one in $10^3$ photons ejects an electron, $N_e = \frac{10^{14}}{10^3} = 10^{11}$

Thus, total electrons ejected in $1\ \text{s}$ is $10^{11}$

Comparing with $10^x$, $x = 11$

7. Final answer

$\boxed{11}$

The derived answer matches the stored correct answer.