Answer
We are between the minima at $m=4$ and $m=5$, so we see (approximately) a maximum.
Work Step by Step
We are trying to determine what appears, on a distant viewing screen, at a point at which the top and bottom rays through the slit have a path length difference equal to $4.5\lambda$.
We use equation (36-3) for a single-slit experiment;
$\alpha \sin \theta = m \lambda $ where $m=1,2,3,....$ (minima)
We find that;
$\alpha \sin \theta = 4.5 \lambda $
$m=4.5$
We are between the 4th and 5th minima, so we are looking at approximately a maximum.
Therefore, we would see a bright spot.
