$\mathbf{\underline{Answer:}\Rightarrow}\;\text{Option}\;3)\;\bbox[lightblue,5px,border: 2px solid blue]{\mathbf{T(n) = \Theta \left (\log^2n.\log\log n\right )}}$
$\mathbf{\underline{Explanation:}\Rightarrow}$
$\mathrm{T(n) = 4T(\sqrt n) + \log ^2 n}$
Let $\mathrm{n=2^m \Rightarrow m = \log_2n}$
$\mathrm{\Rightarrow T(2^m) = 4 T(2^{\frac{m}{2}}) + \log^2 (2^m)}$
$\mathrm{\Rightarrow T(2^m) = 4 T(2^{\frac{m}{2}}) + \log (2^m)\times \log(2^m)}$
Now,
$\mathrm{S(m) =T(2^m) }$
$\mathrm{\Rightarrow S(m) = 4S\left (\dfrac{m}{2}\right) + m\times m}$
$\mathrm{\Rightarrow S(m) = 4S\left (\dfrac{m}{2}\right) + m^2}$
Now, On applying the $\underline{\textbf{Master's Theorem}}$, we get:
$\mathrm{Here, \;a = 4, b = 2, k = 2, p = 0}$
$\therefore \mathrm{a = b^k \Rightarrow 4 = 2^2,\;and\;k = 2, p > 0}$
$\therefore\mathbf{Case\;2\;(i)}$
$\mathrm{T(n) = \Theta(n^{\log_ba}\log^{p+1}n)\;\;,\text{where}\;a = b^k, p>-1}$
$$\therefore\mathrm{T(n) = \Theta(m^{\log_24}\log^{0+1}m)\\=\Theta(m^{\log_22^2}\log^{0+1}m)\\=\Theta(m^{2\log_22}\log^{0+1}m)}$$
$\Rightarrow \mathrm{T(n)= \Theta (m^2\times \log^1m)} \tag{1}$
Now, $\because\mathrm{ m = \log_2n}$
Substituting this value of $\mathbf{m}$ in $(1)$, we get:
$\bbox[lightblue,5px,border: 2px solid blue]{\mathbf{T(n) = \Theta \left (\log^2n.\log\log n\right )}}$
$\therefore \mathrm{option(3)}$ is the right answer.
Refer: Problem $1-2$ Recurrences:
https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-046j-introduction-to-algorithms-sma-5503-fall-2005/assignments/ps1sol.pdf
Master's Theorem Reference: