Answer
$i = 1.0\times 10^{-3}~A$
Work Step by Step
In part (a), we found that $q_0 = 1.0\times 10^{-3}~C$
We can find the current through the resistor at $t= 0$:
$i = (\frac{q_0}{RC})~e^{-t/RC}$
$i = (\frac{q_0}{RC})~e^{-0/RC}$
$i = \frac{q_0}{RC}$
$i = \frac{1.0\times 10^{-3}~C}{(1.0\times 10^6~\Omega)(1.0\times 10^{-6}~F)}$
$i = 1.0\times 10^{-3}~A$
