Answer
(a) less than
(b) $1.2A$ to the left
Work Step by Step
(a) We know that $B\propto \frac{1}{r}$. As the magnitude of the magnetic field by two conductors is equal at point A and as the second wire is closer to point A, hence the magnitude of the current in wire w is less than that in wire 1.
(b) We know that $I_2=\frac{r_2}{r_1}I_1$
We plug in the known values to obtain:
$I_2=\frac{11cm}{33cm}(3.7A)$
$I_2=1.2A$ and it is directed toward the left.
