Answer
The ratio of the diameter of the heaviest string to that of the lightest string is $~~1.02$
Work Step by Step
Let $\rho$ be the density of the material.
We can find an expression for the radius of the heaviest string:
$\rho ~V = M$
$\rho L \pi R^2 = M$
$R^2 = \frac{M}{\rho L \pi}$
$R^2 = \frac{3.0~g/m}{\rho \pi}$
$R = \sqrt{\frac{3.0~g/m}{\rho \pi}}$
We can find an expression for the radius of the lightest string:
$\rho ~V = M$
$\rho L \pi R^2 = M$
$R^2 = \frac{M}{\rho L \pi}$
$R^2 = \frac{2.9~g/m}{\rho \pi}$
$R = \sqrt{\frac{2.9~g/m}{\rho \pi}}$
We can find the ratio of the diameter of the heaviest string to that of the lightest string:
$\frac{2\sqrt{\frac{3.0~g/m}{\rho \pi}}}{2\sqrt{\frac{2.9~g/m}{\rho \pi}}} = \frac{\sqrt{3.0}}{\sqrt{2.9}} = 1.02$
The ratio of the diameter of the heaviest string to that of the lightest string is $~~1.02$
