A function $f$ is asymptotically larger than function $g$ means for large values (all numbers greater than some finite number) $f(n) > g(n).$
$T(n) = 81T\left( \dfrac{n}{9} \right) + 10n^2$ -- This recurrence will asymptotically grow faster than all the other options given.
By applying Master's Theorem'
- $T(n) = 8T \left( \dfrac{n}{4} \right) + 100n^2$
Recursion part, $n^{\log_b a} = n^{\log_4 8}$ and $f(n) = 100 n^2.$ Since, $f(n)$ is polynomially (by a factor of some $n^{\epsilon}, \epsilon > 0)$larger, and $f(n)$ is regular, as per Master Theorem Case 3, $$T(n) = \Theta(f(n))= \Theta(n^2)$$
- $T(n) = 81T \left( \dfrac{n}{9} \right) + 10n^2 \\ \therefore T(n) = O(n^2\log n)$
Recursion part, $n^{\log_b a} = n^{\log_9 81} = n^2$ and $f(n) = 10 n^2.$ Since, $f(n) = \Theta (n^{\log_b a}),$ as per Master Theorem Case 2, $$T(n) = \Theta(f(n) \log n)= \Theta(n^2 \log n)$$
- $T(n) = 16T \left( \dfrac{n}{4} \right) + 100(n\log n)^{1.99}$
Recursion part, $n^{\log_b a} = n^{\log_4 16} = n^2$ and $f(n) = 100 (n \log n)^{1.99}.$ Since, $f(n)$ is polynomially (by a factor of some $n^{\epsilon}, \epsilon > 0)$ smaller, as per Master Theorem Case 1, $$T(n) = \Theta(n^{\log_b a})= \Theta(n^2)$$
- $T(n) = 100T \left( \dfrac{n}{100} \right) + n\log^2n$
Recursion part, $n^{\log_b a} = n^{\log_{100} 100} = n$ and $f(n) = n \log^2 n .$ Here $f(n)$ and $n^{\log_b a}$ are not asymptotically equal and they dont have a polynomial difference. (Any polynomial in $n$ is asymptotically larger than any polynomial in $\log n).$ So Master Theorem cannot be applied as such. But we can apply the Extended Master Theorem Case 2, which states that $$\text{if}\; f(n) = \Theta(n^{\log_b a} \log^k n), k > -1, \text{ then }T(n) = \Theta(n^{\log_b a} \log ^{k+1} n)$$
Here, $k = 2.$ So, we get $$T(n) = \Theta(n \log^3 n)$$
Reference: Master Theorem