Here, $u = 40\,\text{kHz}$, speed of sound in air, $v = 340\,\text{m s}^{-1}$ ∴ $v' =$ speed of bat $= 0.03 \times v$ $= 0.03 \times 340 = 10.20\,\text{m s}^{-1}$The bat moving towards wall acts as a moving source and for the sound waves reflected from the wall, it acts as a moving observer. Thus the source and observer are approaching each other with same speed. i.e. $v_s = v_0 = v' = 10.2\,\text{m s}^{-1}$Thus apparent frequency of the reflected sound waves heard by the bat is given by$\nu' = \frac{v+v_0}{v-v_s} \times \nu = \frac{340+10.2}{340-10.2} \times 40\,\text{kHz}$$= \frac{350.2}{329.8} \times 40 = 42.47\,\text{kHz}$