$\underline{\textbf{Answer:}\Rightarrow}\;\left ( 0.6, 0.4 \right )$
$\underline{\textbf{Solution:}\Rightarrow}$
$\text{Probability of having cavity given the toothache} =\mathrm {P\left ( \dfrac{Cavity}{Toothache} \right ) \\= \dfrac{P\left ( Cavity \wedge Toothache \right )} {P \left (Toothache \right )}\\=\dfrac{\left ( 0.108 + 0.012\right )}{\left ( 0.108+0.012+0.016 +0.064 \right )} \\= \dfrac{0.12}{0.2} \\= 0.6}$
$\mathrm{P\left ( \dfrac{\displaystyle {\neg} Cavity}{Toothache} \right ) = \dfrac{\left ( 0.016 + 0.064\right )}{0.2} = 0.4}$
$\therefore \;\mathbf B$ is the right option.