Answer
(a) $1.5KC$
(b) $8.5 y$
Work Step by Step
(a) We know that
$I=\frac{\Delta Q}{\Delta t}$
This can be rearranged as:
$\Delta Q=I\Delta t$
We plug in the known values to obtain:
$\Delta Q=(0.42A.h)(\frac{3600A.s}{A.h})$
$\Delta Q=1512As$
$\implies \Delta Q=1512C$
$\Delta Q=1.5\times 10^3C=1.5KC$
(b) As $\Delta t=\frac{\Delta Q}{I}$
We plug in the known values to obtain:
$\Delta t=\frac{1512}{5.6\times 10^{-6}}$
$\Delta t=2.7\times 10^8$
$\Delta t=(2.7\times 10^8)(\frac{min}{60})(\frac{h}{60min})(\frac{day}{24h})(\frac{year}{365 days})$
$\Delta t=8.5 y$