Explanation

(d)
Given, $\left| \overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}} \right| = 4$
$\Rightarrow \left| \left| \overrightarrow{\mathbf{a}}\mathbf{\ } \right| \right|\overrightarrow{\mathbf{b}}|\sin{\theta\widehat{\mathbf{n}}\ | = 4}$
$\Rightarrow \left| \left| \overrightarrow{\mathbf{a}}\mathbf{\ } \right| \right|\overrightarrow{\mathbf{b}}\left| \sin{\theta\ | = 4}\ \ \ \lbrack\ \because|\widehat{\mathbf{n}} \right| = 1\rbrack$ ….(i)
Also, $\left| \overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{b}} \right| = 2$
$\Rightarrow \ \left| \left| \overrightarrow{\mathbf{a}}\mathbf{\ } \right| \right|\overrightarrow{\mathbf{b}}|\cos{\theta\ | = 2}$ ….(ii)
On squaring and then on adding Eqs.(i) and (ii), we get
$\left| \overrightarrow{\mathbf{a}} \right|^{2}|\overrightarrow{\mathbf{b}}\left. \ \right|^{2}\sin^{2}{\theta + \left| \overrightarrow{\mathbf{a}} \right|^{2}|\overrightarrow{\mathbf{b}}\left. \ \right|^{2}\cos^{2}{\theta = 4^{2} + 2^{2}}}$
$\Rightarrow \left| \overrightarrow{\mathbf{a}} \right|^{2}|\overrightarrow{\mathbf{b}}\left. \ \right|^{2}(\sin^{2}{\theta + \cos^{2}{\theta) = 16 + 4}}\$
$\Rightarrow \left| \overrightarrow{\mathbf{a}} \right|^{2}|\overrightarrow{\mathbf{b}}\left. \ \right|^{2} = 20$