Answer
$(C)\ 8.5\times 10^{-5}T$
Work Step by Step
We know that
$|\epsilon=N|\frac{(B_2-B_1)\pi r^2}{\Delta t}|$
This can be rearranged as:
$B_2=B_1+\frac{|\epsilon|\Delta t}{N\pi r^2}$
We plug in the known values to obtain:
$B_2=(1.2\times 10^{-5})+\frac{(8.1\times 10^{-4})(0.38)}{3\pi(0.67)^2}=8.5\times 10^{-5}T$