Unless indicated otherwise, assume the speed of sound in air to be v=344ms. Sound is...

a) Our goal is to find the intensity of the sound wave, then to make use of the value of the intensity to determine the energy being delivered to the eardrum each second.
We can calculate the intensity of the sound wave using the relation that describes the intensity level of a sound wave, which is
${\displaystyle \beta =10\log \left(\frac{I}{I_0}\right)}$ (1)
Where ${\displaystyle \beta =20dB,I_0={10}^{-12}\frac{W}{m^2}}$ and I is the intensity of the sound wave that we would like to determine.
Hence
${\displaystyle 20dB=10\log \left(\frac{I}{{10}^{-12}\frac{W}{m^2}}\right)}$
${\displaystyle 2=\log \left(\frac{I}{{10}^{-12}\frac{W}{m^2}}\right)}$
Remember that ${\displaystyle {10}^{\log \left(x\right)}=x}$, hence
${\displaystyle {10}^2={10}^{\log \left(\frac{I}{{10}^{-12}\frac{W}{m^2}}\right)}}$
${\displaystyle \frac{I}{{10}^{-12}\frac{W}{m^2}}={10}^2}$
${\displaystyle I={10}^{-10}\frac{W}{m^2}}$
But we know that intensity is energy per unit time per unit area (Power per unit area), which could be represented mathematically as follows
${\displaystyle E=IA\mathrm{△}}$ (2)
Where A is the area of the tympanic membrane and it can be calculated as follows
${\displaystyle A=\pi r^2=\pi \times {(4.2\times {10}^{-3}m)}^2=5.54\times {10}^{-5}{m}^2}$
Now, substitute into (2) with ${\displaystyle 5.54\times {10}^{-5}{m}^2}$ for ${\displaystyle A,{10}^{-10}\frac{W}{m^2}}$ for I and 1 s for ${\displaystyle \mathrm{△}}$ Notice that ${\displaystyle \mathrm{△}t=1s}$ since we want to calculate the energy is delivered each second
${\displaystyle E=\left({10}^{-10}\frac{W}{m^2}\right)\times (5.54\times {10}^{-5}{m}^2)\times \left(1s\right)}$
${\displaystyle E=5.54\times {10}^{-15}J}$
${\displaystyle l\in e(1,0)\left\{370\right\}}$

b) The velocity of a mosquito with a mass of 2.0 mg and has a kinetic energy of ${\displaystyle 5.54\times {10}^{-15}J}$ can be calculated as follows
${\displaystyle KE=\frac{1}{2}m{v}^2}$
${\displaystyle v=\sqrt{\frac{2\times KE}{m}}}$
To get v, substitute with ${\displaystyle 5.54\times {10}^{-15}J}$ tor KE and ${\displaystyle 2\times {10}^{-6}kg}$
${\displaystyle v=\sqrt{\frac{2\times (5.54\times {10}^{-15}J)}{2\times {10}^{-6}kg}}}$
${\displaystyle v=0.074\times {10}^{-3}\frac{m}{s}}$
${\displaystyle v=0.074\frac{mm}{s}}$