#### Answer

$I = 0.087~kg~m^2$

#### Work Step by Step

We can find the moment of inertia of the lower leg about the knee joint as:
$f = \frac{1}{2\pi}~\sqrt{\frac{Mgd}{I}}$
$2\pi~f = \sqrt{\frac{Mgd}{I}}$
$(2\pi~f)^2 = \frac{Mgd}{I}$
$I = \frac{Mgd}{(2\pi~f)^2}$
$I = \frac{(5.0~kg)(9.80~m/s^2)(0.18~m)}{(2\pi)^2~(1.6~Hz)^2}$
$I = 0.087~kg~m^2$