Option D should be the correct answer.
Consider the following events,
$W$ : India wins,
$W\neg $ : India does not wins (India Lost/ Match Draw/ Match Tie/ Match Suspended etc.)
$X$ : Akbar tells Anthony, "Amar told me that India won"
$X\neg $ : Akbar tells Anthony, "Amar told me that India did not won"
Given $X$, we have to find $W$, that is we have to calculate $P\left(\frac{W}{X}\right)$.
$P\left(\frac{W}{X}\right)$ can be calculated using Bayes's theorem as follow:
$P\left(\dfrac{\text{India Wins}}{\text{Akbar tells Anthony “Amar told me that India won"}}\right)$
$= \dfrac{P\left(\dfrac{\text{Akbar tells Anthony “Amar told me that India won"}}{\text{India Wins}}\right)}{P\left(\frac{\text{Akbar tells Anthony “Amar told me that India won"}}{{\text{India Won}}}\right)\cup P\left(\frac{\text{Akbar tells Anthony “Amar told me that India won"}}{\text{India didn't won}}\right)} $
rewriting same equation using the events defined:
$P\left(\frac{W}{X}\right) = \frac{P\left(\frac{X}{W}\right)}{P\left(\frac{X}{W}\right) + P\left(\frac{X}{W\neg}\right)}\\$
Calculation of $P\left(\frac{X}{W}\right) and \ P\left(\frac{X}{W\neg}\right)\\$ :
$P\left(\frac{X}{W}\right) = P\left(\frac{Case \ 1}{W}\right) \cup \ P\left(\frac{Case \ 4}{W}\right)\\$
$P\left(\frac{Case \ 1}{W}\right) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\\$
$P\left(\frac{Case \ 4}{W}\right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\\$
$So \ P\left(\frac{X}{W}\right) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16}\\$
$P\left(\frac{X}{W\neg}\right) = P\left(\frac{Case \ 6}{W\neg}\right) \cup \ P\left(\frac{Case \ 7}{W\neg}\right)\\$
$P\left(\frac{Case \ 6}{W\neg}\right) = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}\\$
$P\left(\frac{Case \ 7}{W\neg}\right) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}\\$
$So \ P\left(\frac{X}{W\neg}\right) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16}\\$
$Hence, \ P\left(\frac{W}{X}\right) = \frac{\frac{10}{16}}{\frac{10}{16} + \frac{6}{16}} = \frac{10}{16}.\\$
