Moons of Saturn Problem

Lengths scale as the square root of areas.  So, as heureka has calculated,

$${\mathtt{radiusofRhea}} = {\sqrt{{\frac{{\mathtt{34}}}{{\mathtt{3}}}}}}{\mathtt{\,\times\,}}{\mathtt{1\,530}} \Rightarrow {\mathtt{radiusofRhea}} = {\mathtt{5\,150.747\: \!518\: \!564\: \!660\: \!29}}$$

Note that this is 5150km to the nearest ten kilometres.

 $$\boxed{\mbox{Surface area of a sphere: } \quad S=\pi d^2}$$

$$S_{Titan}=\pi d_{Titan}^2$$

$$S_{Rhea}=\pi d_{Rhea}^2$$

$$\dfrac{S_{Titan}}
{S_{Rhea}}
=
\dfrac{\not{\pi} d_{Titan}^2}
{\not{\pi} d_{Rhea}^2}
=\dfrac{34}{3}$$

$$\dfrac{d_{Titan}^2}
{d_{Rhea}^2}
=\dfrac{34}{3}$$

$$d_{Titan}^2=d_{Rhea}^2*\left(\frac{34}{3} \right)$$

$$d_{Titan}=d_{Rhea}*\sqrt{\left(\frac{34}{3} \right)} \quad | \quad d_{Rhea}=1530\;km$$

$$\textstyle{d_{Titan}=1530\;km*\sqrt{\left(\frac{34}{3} \right)} =1530\;km*3.36650164612 = 5150.75\;km\approx5151\;km}$$

Diameter of Titan is 5151 km

Thank you, heureka and Alan.