Answer
(a) $2$
(b) $1020\times 10^{-6}s$
Work Step by Step
(a) We know that $D=\frac{f}{f-number}$. This equation shows that the diameter of the aperture is inversely proportional to the $f-numbers$. The smallest f-number gives the largest aperture diameter and the largest area. As the given camera has $f-stops$ of $2, 2.8, 4, 5.6, 8, 11$ and $16$, hence the f-stop which we can use for the shortest possible exposure time is $2$ because this is the shorter exposure time, which decreases the amount of light entering the camera.
(b) We can find the required shortest exposure time as follows:
$t=(\frac{1}{125}s)(\frac{2}{5.6})^2$
This simplifies to:
$t=1020\times 10^{-6}s$
