Asymptotic estimate for $\sum_{k=1}^{N} {\frac{1}{k^2 H_k}}$
Since the infinite series $\sum_{k=1}^\infty \frac{1}{k^2H_k}$ converges, an asymptotic estimate to your sum is simply $\Theta(1)$.

You can obtain a better estimate by estimating the tail:
$$
\sum_{k=1}^n \frac{1}{k^2H_k} = \sum_{k=1}^\infty \frac{1}{k^2}{H_k} - \sum_{k=n+1}^\infty \frac{1}{k^2H_k}.
$$
We can estimate the tail by
$$
\sum_{k=n+1}^\infty \frac{1}{k^2H_k} \leq
\sum_{k=n+1}^\infty \frac{1}{k(k+1)} = \frac{1}{n},
$$
and so
$$
\sum_{k=1}^n \frac{1}{k^2 H_k} = C - O\left(\frac{1}{n}\right),
$$
where $C = \sum_{k=1}^\infty 1/k^2H_k \approx 1.33275$.