Using Doppler's effect, the frequency observed by the observer when the train moves towards the stationary observer is 

$f_{\mathrm{app}}=f\left(\frac{v}{v-v_{\mathrm{s}}}\right)$

Given here, $v_{\mathrm{s}}=330 \mathrm{m} {\mathrm{s}}^{-1} \mathrm{and} v=66 \mathrm{m} {\mathrm{s}}^{-1}$

Putting the values, 

$f_{\mathrm{app}}=320\left(\frac{330}{330-66}\right)$

$\Rightarrow f_{\mathrm{app}}=400 \mathrm{Hz}$.