Answer
The value of $d$ must exceed $~~72~cm$
Work Step by Step
We can find an expression for $v_b^2$ (the speed squared) after the ball falls to the point directly below the peg:
$\frac{1}{2}mv_b^2 = mgL$
$v_b^2 = 2gL$
In order to swing completely around the peg, the ball must have a minimum speed at the point directly above the peg. This minimum speed is the speed such that the gravitational force provides the centripetal force to keep the ball moving in a circle. We can find an expression for $v_a^2$ (the speed squared) after the ball reaches the point directly above the peg:
$mg = \frac{mv_a^2}{R}$
$v_a^2 = gR$
We can use conservation of energy to find the maximum possible value of $R$:
$\frac{1}{2}mv_b^2 = \frac{1}{2}mv_a^2+mg(2R)$
$v_b^2 = v_a^2+4gR$
$2gL = gR+4gR$
$2L = 5R$
$R = 0.4~L$
Since $R = (L-d)$, then the minimum value of $d$ is $(L-R)$ which is $0.6~L$
We can find the minimum value of $d$:
$d = 0.6~L = (0.6)(120~cm) = 72~cm$
The value of $d$ must exceed $~~72~cm$
