Answer
A capacitor with capacitance $~~36~pF~~$ should be added in parallel.
Work Step by Step
We can find the required ratio $\frac{f_{max}}{f_{min}}$:
$\frac{f_{max}}{f_{min}} = \frac{1.60~MHz}{0.54~MHz} = 2.963$
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 find an expression for the minimum frequency $f_{min}$:
$\omega = \frac{1}{\sqrt{L~C}}$
$2\pi~f_{min} = \frac{1}{\sqrt{L~C_{max}}}$
$f_{min} = \frac{1}{2\pi~\sqrt{L~C_{max}}}$
When a capacitor of capacitance $C$ is added in parallel, then $C_{min} = C+10~pF$ and $C_{max} = C+365~pF$
We can find $C$:
$\frac{f_{max}}{f_{min}} = \frac{\frac{1}{2\pi~\sqrt{L~C_{min}}}}{\frac{1}{2\pi~\sqrt{L~C_{max}}}}$
$\frac{f_{max}}{f_{min}} = \frac{2\pi~\sqrt{L~C_{max}}}{2\pi~\sqrt{L~C_{min}}}$
$\frac{f_{max}}{f_{min}} = \frac{\sqrt{C_{max}}}{\sqrt{C_{min}}}$
$\frac{f_{max}}{f_{min}} = \sqrt{\frac{C_{max}}{C_{min}}} = 2.963$
$\frac{C_{max}}{C_{min}} = 8.779$
$C_{max} = 8.779~C_{min}$
$C+365~pF = 8.779~(C+10~pF)$
$7.779~C = 277.21~pF$
$C = \frac{277.21~pF}{7.779}$
$C = 36~pF$
A capacitor with capacitance $~~36~pF~~$ should be added in parallel.
