Step 1: Note what a radius must look like dimensionally.
$R$ is a length, so whatever combination of $r_1$ and $r_2$, which are also lengths, gives $R$ must itself reduce to a single power of length, not length squared or length cubed.
Step 2: Check each option for correct dimensions.
$r_1 r_2^2$ and $r_1^2 r_2$, taken directly with no root, both have dimension length cubed, since they multiply three lengths together. Neither can be a radius on its own, so options (B) and (D) are ruled out right away.
$(r_1 r_2^2)^{2/3}$ starts as length cubed and is then raised to the power $2/3$, giving dimension length squared, again not a valid radius. This rules out option (A).
$(r_1^2 r_2)^{1/3}$ starts as length cubed and is raised to the power $1/3$, giving exactly dimension length, which is what a radius should be. Option (C) is the only dimensionally valid choice.
Step 3: Confirm with the volume argument.
This dimensional shortcut matches the actual physics too. The equivalent sphere is defined to have the same volume as the ellipsoid, whose volume is $\frac{4}{3}\pi r_1^2 r_2$ (two equatorial semi axes and one polar semi axis). Setting this equal to a sphere's volume $\frac{4}{3}\pi R^3$ and solving gives $R = (r_1^2 r_2)^{1/3}$, the same result.
Step 4: Final Answer.
$R = (r_1^2 r_2)^{1/3}$, so the correct option is (C).
\[ \boxed{R = \left(r_1^2 r_2\right)^{1/3}} \]