Answer
$L=5.0\times 10^{-21}H$
Work Step by Step
We know that;
$\omega=\frac{1}{\sqrt{LC}}$
This can be rearranged as:
$L=\frac{1}{\omega ^2 C}$..........eq(1)
Next, we find the frequency $f$;
$f=\frac{c}{\lambda}=\frac{3\times 10^8}{550\times 10^{-9}}=5.4545\times 10^{14}Hz$
This enables us to find $w$:
$\omega=2\pi f$
$\omega= 2(3.1416)(5.4545\times 10^{14})=3.4272\times 10^{15}\frac{rad}{s}$
We plug in the known values in eq(1)to obtain:
$L=\frac{1}{(3.4272\times 10^{15})^2(17\times 10^{-12})}=5.0\times 10^{-21}H$
