Answer
The free-fall acceleration on Mars is $3.67~m/s^2$
Work Step by Step
We can write an expression for the period of the pendulum on the earth as:
$T_e = 2\pi~\sqrt{\frac{L}{g}}$
We can write an expression for the period of the pendulum on Mars as:
$T_m = 2\pi~\sqrt{\frac{L}{g_m}}$
We can divide the first expression by the second expression.
$\frac{T_e}{T_m} = \frac{2\pi~\sqrt{\frac{L}{g}}}{2\pi~\sqrt{\frac{L}{g_m}}}$
$\frac{T_e}{T_m} = \sqrt{\frac{g_m}{g}}$
$g_m = (\frac{T_e}{T_m})^2~g$
$g_m = (\frac{1.50~s}{2.45~s})^2~(9.80~m/s^2)$
$g_m = 3.67~m/s^2$
The free-fall acceleration on Mars is $3.67~m/s^2$.