In Doppler’s formula, $ \displaystyle \nu =\frac{v+{{v}_{l}}}{v-{{v}_{s}}}{{\nu }_{0}}$ the speeds $v_s$ and $v_l$ both are with respect to the medium.

So, $\displaystyle {{\text{v}}_{s}}\text{=0-10=-10}$
$\displaystyle {{\text{v}}_{l}}\text{=10-10=0}$

So, $\displaystyle \nu =\frac{340+0}{340-(-10)}\times 600=583Hz$