Answer
$(\frac{C}{A})^2I_{AB}$
Work Step by Step
We know that
$V=I_{AB}R_{AB}$
$\implies V=I_{AB}\rho\frac{L}{A}$
$\implies V=I_{AB}\rho\frac{C}{AB}$......(1)
Now, if the potential difference V is maintained between the two faces $BC$ of the block, then
$V=I_{BC}R_{BC}$
$\implies V=I_{BC}\rho \frac{L}{A}$
$\implies V=I_{BC}\rho \frac{A}{BC}$....... (2)
Comparing eq (1) and eq(2), we obtain:
$I_{AB}\rho (\frac{C}{AB})=I_{BC}\rho (\frac{A}{BC})$
$\implies I_{BC}=I_{AB}(\frac{C}{A})^2$
