Answer
$B = 3.3~\mu T$
Work Step by Step
We found that the magnitude of the magnetic field in region 1 is $B = 10~\mu T$ and the direction is out of the page.
As the loop leaves region 1, there will be an induced current due to the change in flux from the magnetic field in region 1. We can find the magnitude of this induced current:
$i = \frac{BHv}{R}$
$i = \frac{(10\times 10^{-6}~H)(0.015~m)(0.40~m/s)}{0.020~\Omega}$
$i = 3.0~\mu A$
Since the magnetic flux from the magnetic field in region 1 is decreasing, the current in the loop is counterclockwise.
From the graph, we can see that the net induced current in the loop is $2.0~\mu A$ in the clockwise direction.
Thus the induced current due to the change in magnetic flux from the magnetic field in region 2 is $1.0~\mu A$ clockwise.
We can find the magnitude of the magnetic field in region 2:
$i = \frac{BHv}{R}$
$B = \frac{iR}{Hv}$
$B = \frac{(1.0\times 10^{-6}~A)(0.020~\Omega)}{(0.015~m)(0.40~m/s)}$
$B = 3.3~\mu T$
