#### Answer

Please see the work below.

#### Work Step by Step

We know that the speed of the bug at a distance of y from the top is given as
$mgy=(\frac{1}{2})mv^2$
This simplifies to:
$v=\sqrt{2gy}$
We also know that the centripetal force is given as:
$mgcos(\theta)=mv^2R$
$\implies mg(\frac{y}{R})=mv^2R$
$\implies mg(\frac{y}{R})=m(\sqrt{2gy})^2{R}$
$mg(\frac{R-y}{R})=m(\frac{2gy}{R})$
$R-y=2y$
$y=\frac{R}{3}$