The self-inductance of a solenoid is given by:
[tex]L= \frac{\mu_0 N^2 A}{l} [/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid
For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is
[tex]A= \frac{Ll}{\mu N^2}= \frac{(0.025 H)(0.70 m)}{(4\pi \cdot 10^{-7}N/A^2)(3000)^2}= 1.55 \cdot 10^{-3}m^2[/tex]
And since the area is related to the radius by
[tex]A=\pi r^2[/tex]
The radius of the solenoid is
[tex]r= \sqrt{ \frac{A}{\pi} } = \sqrt{ \frac{1.55 \cdot 10^{-3} m^2}{\pi} } =0.022 m=2.2 cm[/tex]
