Answer
$ 6.1\times 10^7\;\rm ion$
Work Step by Step
As the author told us, we can consider the cell's membrane as a parallel plate capacitor. And let's assume that the cell is shaped as a sphere.
Recalling that the potential difference of a parallel plate capacitor is given by
$$\Delta V_C=\dfrac{Q}{C}$$
So the charge in potential difference requires a change in the charge, so
$$\Delta Q=C\Delta V_C$$
$$\Delta Q=C(V_f-V_i)$$
where $\Delta Q$ is the net charge of the sodium ions that enter the cell. The charge of one sodium ion equals the charge of one proton $q=e$.
Hence,
$$Ne=C(V_f-V_i)$$
where $N$ is the number of sodium ions;
$$N=\dfrac{C(V_f-V_i)}{e}$$
where $C=\dfrac{\epsilon_0\kappa A}{d}$
$$N=\dfrac{\epsilon_0\kappa A(V_f-V_i)}{ed}$$
and $A=4\pi R^2$
Plug the known;
$$N=\dfrac{4\pi \epsilon_0\kappa AR^2(V_f-V_i)}{ed}$$
$$N=\dfrac{4\pi (8.85\times 10^{-12})(9) (25\times 10^{-6})^2(40-(-70))}{(1.6\times 10^{-19})(7\times 10^{-6})}$$
$$N=\color{red}{\bf 6.1\times 10^7}\;\rm ion$$
