Square roots can be broken up (ONLY if there is a single term under the root, which there is in this case).
\(\displaystyle \sqrt{100x^2y^{16}z^8} = \sqrt{10^2} \sqrt{x^2} \sqrt{y^{16}} \sqrt{z^8}\)
Then, you can break higher powers into squares:
\(\displaystyle \sqrt{10^2} \sqrt{x^2} \sqrt{y^{16}} \sqrt{z^8} = \sqrt{10^2} \sqrt{x^2} \sqrt{(y^{8})^2} \sqrt{(z^4)^2}\)
Now you can cancel call the squares with the square roots:
\(\displaystyle \sqrt{10^2} = 10\\ \sqrt{x^2} = |x| \text{ (if you are assuming x is positive you may omit the absolute value)} \\ \sqrt{(y^{8})^2} = y^8 \\ \sqrt{(z^4)^2} = z^4\)
So,
\(\displaystyle \sqrt{10^2} \sqrt{x^2} \sqrt{(y^{8})^2} \sqrt{(z^4)^2} = 10*|x|y^8z^4\)