#### Answer

The length of the pendulum is 0.33 meters.

#### Work Step by Step

We can find the period of a 2.0-meter long pendulum on the earth as:
$T = 2\pi~\sqrt{\frac{L}{g}}$
$T = 2\pi~\sqrt{\frac{2.0~m}{9.80~m/s^2}}$
$T = 2.84~s$
We can find the length $L_m$ of a pendulum that would have the same period on the moon, which has an acceleration due to gravity of $g_m = 1.63~m/s^2$. Therefore;
$T = 2\pi~\sqrt{\frac{L_m}{g_m}}$
$L_m = \frac{T^2~g_m}{(2\pi)^2}$
$L_m = \frac{(2.84~s)^2(1.63~m/s^2)}{(2\pi)^2}$
$L_m = 0.33~m$
The length of the pendulum is 0.33 meters.