When both \( r_1 \) and \( r_2 \) are greater than \( R \) (i.e., both are outside the sphere), the gravitational force at a distance \( r \) from the center of a uniform sphere is given by:

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

So, the forces \( F_1 \) and \( F_2 \) at distances \( r_1 \) and \( r_2 \) from the center are:

\[
F_1 = \frac{G M}{r_1^2}
\]
\[
F_2 = \frac{G M}{r_2^2}
\]

Now, taking the ratio \( \frac{F_1}{F_2} \):

\[
\frac{F_1}{F_2} = \frac{\frac{G M}{r_1^2}}{\frac{G M}{r_2^2}} = \frac{r_2^2}{r_1^2}
\]

Thus, the ratio of the gravitational forces is:

\[
\frac{F_1}{F_2} = \left( \frac{r_2}{r_1} \right)^2
\]

To find the dimensions of the gravitational constant \( G \), use Newton's law of gravitation:

\[
F = \frac{G M_1 M_2}{r^2}
\]

Where:
- \( F \) is force (with dimensions \( [M L T^{-2}] \)),
- \( M_1 \) and \( M_2 \) are masses (with dimensions \( [M] \)),
- \( r \) is distance (with dimensions \( [L] \)).

Rearranging for \( G \):

\[
G = \frac{F r^2}{M_1 M_2}
\]

Substitute the dimensions:

\[
G = \frac{[M L T^{-2}] [L^2]}{[M][M]}
\]

Simplify:

\[
G = [M^{-1} L^3 T^{-2}]
\]

Thus, the dimensions of \( G \) are \( [M^{-1} L^3 T^{-2}] \).