Answer
$2.43×10^{-14}\;\rm J$
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.
We know that the energy stored in a capacitor is given by
$$U=V_Lu_e$$
where $V_L$ is the volume and $u$ is the volume energy density stored in the electric field; where $u=\epsilon_0(\Delta V/d)^2/2$
$$U=V_L \frac{\epsilon_0}{2}\left(\dfrac{\Delta V}{d}\right)^2$$
where $V_L$ is the area of the membrane, see the figure below, which is given by $A=\frac{4}{3}\pi (R+d)^3-\frac{4}{3}\pi R^3$
$$U=\frac{4}{3}\pi \left[(R+d)^3- R^3\right]\frac{\epsilon_0}{2}\left(\dfrac{\Delta V}{d}\right)^2$$
$$U=\frac{2}{3}\pi \epsilon_0\left[(R+d)^3- R^3\right] \left(\dfrac{\Delta V}{d}\right)^2$$
Plug the known;
$$U=\frac{2}{3}\pi (8.85\times 10^{-12})\left[(25\times 10^{-6}+7\times 10^{-9})^3- (25\times 10^{-6})^3\right] \left(\dfrac{-70\times 10^{-3}}{7\times 10^{-9}}\right)^2$$
$$U=\color{red}{\bf 2.43×10^{-14}}\;\rm J$$
