Solution:

The intersection of the plane on which the base of the pyramid lies with the sphere in which it is inscribed is a circle to which the points $A, B, C, D$ belong. Consequently the quadrilateral $ABCD$ is inscribable in a circle and therefore has opposite angles supplementary (because two by two they subtend supplementary arcs of the same circle).

Let us now consider the triangles $ADE$ and $CBE$. They have the angle at $E$ in common, moreover $E \hat{C} B = E \hat{A} D$ because both are supplementary to $B \hat{C} D$. Similarly $E \hat{B} C = E \hat{D} A$ because both are supplementary to $A \hat{B} C$. Therefore the triangle $ADE$ is similar to $CBE$ (having their angles respectively equal) with ratio of similarity $2:1$.

Consequently the area of $ADE$ is four times the area of $BCE$ and $4/3$ of the area of $ABCD$. Since the heights of the two given pyramids coincide (both the point $V$ and the base planes coincide), the ratio between the volumes of the pyramids $VABCD$ and $VADE$ coincides with the ratio between the base areas $ABCD$ and $ADE$, and is therefore $4/3$.