Interpretation of “number of lines of force per unit surface area”

In electrostatics, the number of electric lines of force crossing per unit area is proportional to the electric field magnitude.

So we need the electric field just outside the surface of the uniformly charged sphere: $E=\frac{Q}{4\pi \epsilon_0 R^2}$

For a uniformly charged sphere, $Q=\rho \cdot \frac{4}{3}\pi R^3$ Hence, $E=\frac{\rho R}{3\epsilon_0}$

Convert charge density into SI units

Given: $\rho=2\,\mu C\,cm^{-3}$

Now, $1\,\mu C=10^{-6}\,C, \qquad 1\,cm^3=10^{-6}\,m^3$

Therefore, $1\,\mu C\,cm^{-3}=\frac{10^{-6}C}{10^{-6}m^3}=1\,C\,m^{-3}$

So, $\rho=2\,C\,m^{-3}$

Substitute values

Radius: $R=6\,m$

Using $E=\frac{\rho R}{3\epsilon_0}$ we get $E=\frac{2\times 6}{3\times 8.85\times 10^{-12}}$

$E=\frac{12}{26.55\times 10^{-12}}$

$E\approx 0.452\times 10^{12}$

$E\approx 4.52\times 10^{10}\,N C^{-1}$

Match with the required form

Required form is: $\underline{\hspace{1cm}}\times 10^{10}\,NC^{-1}$

Thus, $E\approx 4.52\times 10^{10}\,NC^{-1}$

The integer asked is therefore approximately $5$

Comparison with stored answer

My derived answer is $5$, while the stored correct answer is $45$.

The value $45$ would correspond to writing $4.5\times 10^{10}$ and then mistakenly taking the coefficient as $45$ instead of $4.5$.