Answer
6.92 mA/s.
Work Step by Step
By definition, $L=\frac{N\Phi_B}{i}$.
$$ L=\frac{800(3.25\times10^{-3}Wb) }{2.90A}=0.8966H$$
$$L = \frac{\epsilon}{|di/dt|}$$
Solve for the rate of change of the current.
$$|\frac{di}{dt}|=\frac{|\epsilon|}{L}=\frac{6.20\times10^{-3}V}{0.8966H}=6.92 mA/s$$