Consider a solid cube of uniform volume charge density. If V is electrostatic potential at one of the corners of cube then the potential at the centre of cube is

Let  $\rho$ is the charge density and $l$ is the edge length of cube. We know that potential at any point has to be proportional to charge and inversely proportional to side length of cube so
$V\propto \frac{Q}{l}=\frac{\rho l^3}{l}=\rho l^2$ 
Now, to find potential at centre, the cube of edge length $l$  can be considered as 8 cubes of length $l/2$ so the potential at centre of original can be obtained as sum of potentials due to 8 cubes (of length $l/2$ ) at the corners
$$\begin{gathered} V_{centre}\propto 8\times \rho {\left(\frac{l}{2}\right)}^2=\frac{8\rho l^2}{4}=2\rho l^2 \\ \therefore \frac{V_{centre}}{V}=\frac{2\rho l^2}{\rho l^2}=2 \\ \Rightarrow V_{centre}=2V \end{gathered}$$