Answer
We can see that the decrease in mechanical energy is $~~-2mgL$
Work Step by Step
In part (a), we found that $v_0= \sqrt{2~g~L}$
We can use conservation of energy to find the speed $v_B$ at point $B$ when there is no friction:
$\frac{1}{2}mv_B^2 = \frac{1}{2}mv_0^2+mgL$
$v_B^2 = v_0^2+2gL$
$v_B^2 = (\sqrt{2gL})^2+2gL$
$v_B^2 = 4gL$
$v_B = \sqrt{4gL}$
We can find the mechanical energy at point B when there was no friction:
$E = K_B+U_B$
$E = \frac{1}{2}mv_B^2+0$
$E = \frac{1}{2}m(\sqrt{4~g~L})^2$
$E = 2mgL$
We can find the mechanical energy at point B when the ball is at rest:
$E = K_B+U_B$
$E = 0+0$
$E = 0$
We can see that the decrease in mechanical energy is $~~-2mgL$
