Answer
$L = 2.2\times 10^{-4}~H$
Work Step by Step
In part (b), we found that a capacitor with capacitance $~~36~pF~~$ should be added in parallel.
Then the minimum capacitance is $10~pF+36~pF = 46~pF$
We can find an expression for the maximum frequency $f_{max}$:
$\omega = \frac{1}{\sqrt{L~C}}$
$2\pi~f_{max} = \frac{1}{\sqrt{L~C_{min}}}$
$f_{max} = \frac{1}{2\pi~\sqrt{L~C_{min}}}$
We can use the expression for the maximum frequency $f_{max}$ to find $L$:
$f_{max} = \frac{1}{2\pi~\sqrt{L~C_{min}}}$
$f_{max}^2 = \frac{1}{4\pi^2~L~C_{min}}$
$L = \frac{1}{4\pi^2~C_{min}~f_{max}^2}$
$L = \frac{1}{(4\pi^2)~(46\times 10^{-12}~F)~(1.60\times 10^6~Hz)^2}$
$L = 2.2\times 10^{-4}~H$
