Answer
The beat frequency is $3.2~Hz$
Work Step by Step
We can find the frequency received at the stationary platform from the tuba on the approaching train.
$f' = \frac{f}{(1-\frac{v_{source}}{v_{snd}})}$
$f' = \frac{75~Hz}{(1-\frac{14.0~m/s}{343~m/s}~)}$
$f' = 78.2~f$
The beat frequency is the difference between the two frequencies $75~Hz$ and $78.2~Hz$. Therefore, the beat frequency is $3.2~Hz$.
