Let $\sigma(n)$ be the number of positive divisors of $n$, and let $\operatorname{rad} n$ be the product of the distinct prime divisors of $n$. By convention, $\operatorname{rad} 1 = 1$. Find the greatest integer not exceeding 
\[ 100\left(\sum_{n=1}^{\infty}\frac{\sigma(n)\sigma(n \operatorname{rad} n)}{n^2\sigma(\operatorname{rad} n)}\right)^{\frac{1}{3}}. \]

Let's think step by step and output the final answer within \boxed{}.
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