Consider a very rough model of a beryllium atom. The nucleus...

To find the electric field at a point just inside the $2s$ cloud, we construct a spherical Gaussian surface of radius $r = 5.2 \times 10^{-11} \text{ m}$ centered at the nucleus.

The charge enclosed by this Gaussian surface includes the nucleus and the two $1s$ electrons. The $2s$ electrons lie outside this surface and do not contribute to the enclosed charge.

Charge of the nucleus, $q_n = +4e$
Charge of the two $1s$ electrons, $q_{1s} = -2e$

Net enclosed charge, $q_{enc} = +4e - 2e = +2e$

Using Gauss's law, the electric field $E$ at this distance is given by:

$E = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_{enc}}{r^2}$

Substituting the given values:

$E = \dfrac{9 \times 10^9 \times 2 \times 1.6 \times 10^{-19}}{(5.2 \times 10^{-11})^2}$

$E = \dfrac{28.8 \times 10^{-10}}{27.04 \times 10^{-22}}$

$E \approx 1.065 \times 10^{12} \text{ N C}^{-1}$

Rounding to two significant figures, we get $E \approx 1.1 \times 10^{12} \text{ N C}^{-1}$.