If  $$f$$  is the frequency and  $$lambda$$  is the wavelength of an electromagnetic wave, then  $$flambda =c$$. The frequency is the same as the frequency of oscillation of the current in the  $$LC$$  circuit of the generator. That is, $$f=1/2pi sqrt{LC}$$, where  $$C$$  is the capacitance and  $$L$$  is the inductance. Thus, $$ rac{lambda}{2pi sqrt{LC}}=c$$. The solution for  $$L$$  is, $$L=dfrac{lambda ^{2}}{4pi ^{2}Cc^{2}}=dfrac{left ( 550*10^{-9}m \right )^{2}}{4pi ^{2}left ( 17*10^{-2}F \right )left ( 2.998*10^{8}m/s \right )^{2}}=5.00*10^{-21}H$$. This is exceedingly small.