Answer
$a = 3.9 ~m/s^2$
Work Step by Step
We can find the velocity of the speck as it completes 45 rpm on the wheel through the formula:
$v = \frac{(45)(2\pi r)}{T}$
$v = \frac{(45)(2\pi)(0.175 ~m)}{60 ~s}$
$v = 0.825~m/s$
We can then use the velocity of the speck to find the acceleration:
$a = \frac{v^2}{r}$
$a = \frac{(0.825 ~m/s)^2}{(0.175 ~m)}$
$a = 3.9 ~m/s^2$