How to determine the orbital period of an asteroid using Kepler laws given its orbit's radius?
Here's how I would do it. I'd convert everything to a single, standard set of units as recommended in the comments, and also stick to one digit before the decimal in scientific notation:

$$ T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}$$

Using all numbers in the same units:

$r \ 4.5 \times 10^{11} \ (m) $

$G = 6.674 \times 10^{-11} \ (m^3 \ kg^{-1}  s^{-2}) $

$M = 1.989 \times 10^{30} \ (kg)$

$$ T^2=\frac{4 \cdot \pi^2 \cdot 9.11 \times 10^{34}}{6.674 \times 10^{-11} \cdot 1.989 \times 10^{30}} = \frac{3.60 \times 10^{36}}{1.33 \times 10^{20}} = 2.71 \times 10^{16} (s)$$

$T = 1.65 \times 10^8 \ (s) $

$\frac{1.65 \times 10^8}{365.25 \times 24 \times 3600} = 5.21 \ (years)$



Just fyi you can also call the product $GM$ the standard gravitational parameter and those numbers are more accurate than $G$ and $M$ are separately, as discussed in this answer.

The comments under your question also mention that it is sometimes easier to use AU and years as units and use Earth's orbit for scaling to the 3/2 power, but only for things orbiting our sun.  With $r = 3$ AU:

$T^2 = 3^3$

$T = 3^{3/2} = 5.2 \ (years)$