Answer
$64$
Work Step by Step
Let $C_{m}=256$ denote the frequency of middle C.
and $C_{m\pm N}$ denote frequencies N octaves above/below middle C.
We are given $C_{m+1}=2\cdot C_{m}$
Since $\{C_{n}\}$ is a geometric sequence, the common ratio is
$r=\displaystyle \frac{C_{m+1}}{C_{m}}=\frac{512}{256}=2$
A part of the sequence near $C_{m}$:
$... C_{m-2}, C_{m-1}, C_{m}, C_{m+1}..., $
is such that
$C_{m}=C_{m-1}\cdot r=C_{m-2}\cdot r^{2}$
So,
$C_{m-2}=\displaystyle \frac{C_{m}}{r^{2}}=\frac{256}{4}=64$
