A more formal approach:
$f(n) = n^2 \log n$
$g(n) = n (\log n)^{10}$
We can use the limit definition of $O$-notation
http://cse.unl.edu/~choueiry/S06-235/files/Asymptotics-HandoutNoNotes.pdf
$\displaystyle{\lim_{n \to \infty }} \frac{f(n)}{g(n)} = 0, \implies f(n) = o(g(n))$
small $o$ implying $f$ is strictly asymptotically lower than $g$. Also by definition, $o \implies O \text{ but } O \not\implies o$.
$\displaystyle{\lim_{n\to \infty}} \frac{f(n)}{g(n)} = c, c > 0 \implies f(n) = \Theta(g(n))$
$\displaystyle{\lim_{n \to \infty}} \frac{f(n)}{g(n)} = \infty, \implies f(n) = \omega(g(n))$
small $\omega$ implying $f$ is strictly asymptotically higher than $g$. Also by definition, $\omega \implies \Omega, \text{ but } \Omega \not \implies \omega.$
We can use this to prove the above question
For any $k > 0$
$\displaystyle{\lim_{n \to \infty}} \frac{(\log n)^{k}}{ n}$
Applying L'Hôpital's rule,
$\implies \displaystyle{ \lim_{n \to \infty}} \frac{ k*(\log n)^{k-1}}{ n}$
$= \displaystyle{\lim_{n \to \infty}} \frac{ k!}{n} = 0$
So, $(\log n)^k = o(n)$
Now for large $n$, $n >> (\log n)^9$
i.e., $n^{2}\log n >> n(\log n)^{10}$ (Multiplying both sides by $n \log n)$
So, $n(\log n )^{10} = o(n^{2}\log n)$ and $n(\log n )^{10} \neq \Theta(n^{2}\log n)$ (making LHS strictly asymptotically lower than RHS)
Or
$n(\log n )^{10} = O(n^{2}\log n)$ and $n^{2}\log n \neq O(n(\log n )^{10})$
Option $B$.