Answer
$a = 8.0~mV/m^2$
Work Step by Step
Note that $a = \frac{B}{t}$, so $a$ is the slope of the $B$ versus $t$ graph.
Since the current drops to $\frac{1}{3}$ of its value between $t = 20~s$ and $t = 30~s$, the induced emf in the loop must be $\frac{2}{3}$ of the battery emf and in the opposite direction. Thus the induced emf is $4.00~\mu V$
We can use the magnitude of the induced emf to find $a$:
$emf = \frac{d\Phi}{dt}$
$emf = A~\frac{dB}{dt}$
$emf = A~a$
$a = \frac{emf}{A}$
$a = \frac{4.00\times 10^{-6}~V}{5.0\times 10^{-4}~m^2}$
$a = 8.0~mV/m^2$
