The inductance of a solenoid is given by
[tex]L=\mu \frac{N^2}{l} A[/tex]
where
[tex]\mu = 12.56 \cdot 10^{-7}NA^{-2}[/tex] is the magnetic permittivity
N is the number of turns of the solenoid
l is its lenght
A is its cross-sectional area
By re-arranging the formula and replacing the numbers, we get the number of turns of the solenoid:
[tex]N= \sqrt{ \frac{lL}{\mu A} }= \sqrt{ \frac{(4.7\cdot 10^{-3}H)(0.3 m)}{(12.56\cdot 10^{-7}NA^{-2})(3.3\cdot10^{-2}m^2)} } =583[/tex]
