Solution:

The initial gravitational force between two bodies is:

\[
F = G \frac{m \cdot m}{r^2}
\]

After transferring 25% of mass from one body to the other, the new masses become \( 0.75m \) and \( 1.25m \), and the separation becomes \( \frac{r}{2} \). The new gravitational force is:

\[
F' = G \frac{(0.75m)(1.25m)}{\left(\frac{r}{2}\right)^2}
\]

Simplifying:

\[
F' = G \frac{0.9375 m^2}{\frac{r^2}{4}} = 4 \times 0.9375 \times \frac{G m^2}{r^2} = 15F/4
\]

So, the new gravitational force is:

\[
F' = 3.75F= 15F/4
\]