A cube of side b has a charge q at seven of its vertices. The electric field due to this charge distribution at the centre of this cube will be:

If same charge is placed at all eight vertices then net electric field is zero i.e., vector sum of all eight fields is zero.
$$\begin{gathered} {\overrightarrow{\mathrm{E}}}_1+{\overrightarrow{\mathrm{E}}}_2+\overrightarrow{{\mathrm{E}}_3}+\overrightarrow{{\mathrm{E}}_4}+\overrightarrow{{\mathrm{E}}_5}+\overrightarrow{{\mathrm{E}}_6}+\overrightarrow{{\mathrm{E}}_7}+\overrightarrow{{\mathrm{E}}_8}=0 \\ \left|\overrightarrow{{\mathrm{E}}_1}+{\overrightarrow{\mathrm{E}}}_2+{\overrightarrow{\mathrm{E}}}_3+{\overrightarrow{\mathrm{E}}}_4+{\overrightarrow{\mathrm{E}}}_5+{\overrightarrow{\mathrm{E}}}_6+{\overrightarrow{\mathrm{E}}}_7\right|=\left|{\overrightarrow{\mathrm{E}}}_8\right|=\frac{4\mathrm{k}\cdot \mathrm{q}}{3{\mathrm{b}}^2} \end{gathered}$$
As half of the diagonal of the cube is $\mathrm{r}=\frac{\sqrt{3}\mathrm{b}}{2}$
Hence when seven charges are at vertices then electric field at the center $=\frac{4\mathrm{k}\cdot \mathrm{q}}{3{\mathrm{b}}^2}$