Answer
For note A
$27.5Hz;12.5m$
For note C
$4.19KHZ;8.18\times 10^{-2}m$
Work Step by Step
We can find the required frequencies and wavelengths as follows:
The lowest note on the piano is A and the frequency given in the table=440Hz
We are given that note A has octaves below 440 Hz
Hence $f=\frac{440}{16}=27.5Hz$
We know that
$\lambda=\frac{v}{f}$
We plug in the known values to obtain:
$\lambda=\frac{343m/s}{27.5Hz}$
$\lambda=12.5m$
Now the frequency of note C is given as
$f^{\prime}=261.7Hz\times 16$
$f^{\prime}=4.19KHz$
and $\lambda^{\prime}=\frac{343}m/s{4.19KHz}=8.18\times 10^{-2}m$