Answer
As the bob passes the lowest point, the tension in the cord is $2mg$
Work Step by Step
We can use conservation of energy to find the speed at the lowest point:
$KE = U_g$
$\frac{1}{2}mv^2 = mg~(\frac{L}{2})$
$v = \sqrt{gL}$
The net force at the bottom provides the centripetal force to keep the bob moving in a circular path. We can find the tension $F_T$ in the cord:
$\sum F = \frac{mv^2}{L}$
$F_T-mg = \frac{mv^2}{L}$
$F_T = \frac{m(\sqrt{gL})^2}{L}+mg$
$F_T = \frac{mgL}{L}+mg$
$F_T = 2mg$
As the bob passes the lowest point, the tension in the cord is $2mg$
