Answer
For the $12^{th}$ harmonics of B, the frequency match the frequency of A’s third harmonic. The harmonic does not belong to the the first eight harmonics of string B.
Work Step by Step
For a stretched string of length $L$ with fixed ends, the resonant frequencies are
$f=\frac{v}{\lambda}$
or, $f=\frac{n}{2L}\sqrt {\frac{T}{\mu}}\;\;\;\text{for}\;n=1, 2, 3, . . . .$
For string A
$f_A=\frac{n}{2L}\sqrt {\frac{T}{\mu}}$
For string B
$f_B=\frac{n}{2\times(4L)}\sqrt {\frac{T}{\mu}}$
Let, For the $n^{th}$ harmonics of B, the frequency match the frequency of A’s third harmonic. Therefore,
$\frac{n}{2\times(4L)}\sqrt {\frac{T}{\mu}}=\frac{3}{2L}\sqrt {\frac{T}{\mu}}$
or, $n=12$
Therefore, for the $12^{th}$ harmonics of B, the frequency match the frequency of A’s third harmonic. The harmonic does not belong to the the first eight harmonics of string B.
