A pendulum is carried to the moon where gravity is 1/6th the value on earth. will the clock run faster/slower?

If you look at the equation describing the period of the pendulum, the period is inversely proportional to the square root of the acceleration due to gravity. If the acceleration due to gravity is reduced by a factor of 6, then as shown in your equation (t=2pi*sqrt(6L/g)), the period of the pendulum on the moon is sqrt(6) times the period of the same pendulum on the earth. Hence the clock runs slower by a factor of sqrt(6)