Answer
(b) $f_{s,A}=f_{s,B}=\frac{1}{2}f_{s,C}$
Work Step by Step
Let, the masses of each block is $m$.
According to the given diagram,
The static friction on C arises only from the contact with B
$f_{s,A}=\mu mg$, directed towards right
The static friction on B arises from the contact with C and A
The static friction on B arises at the contact with A: $f^{1}_{s,B}=\mu mg$, directed towards left
The static friction on B arises at the contact with C: $f^{2}_{s,B}=\mu (m+m)g$, directed towards right
Therefore resultant static friction on B
$f_{s,B}=f^{2}_{s,B}-f^{1}_{s,B}=\mu mg$, directed towards right.
The static friction on A arises only from the contact with B
$f_{s,C}=\mu (m+m)g=2\mu mg$, directed towards left
Thus, $f_{s,A}=f_{s,B}=\frac{1}{2}f_{s,C}$
