If A throw first then A's chances of winning =$\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+...$
$=1/6\left(1+\left(\frac{5}{6}\right)^2+(\frac{5}{6}.\frac{5}{6})^2+ \dots \right)$
$=1/6\left(\frac{1}{1-(5/6)^2}\right)=6/11$
And B's chances of winning = $\frac{5}{6}.\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}+\dots$
$=\frac{5}{6}.\frac{1}{6}\left(1+\frac{5}{6}.\frac{5}{6}+\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{5}{6}+\dots \right)$
$=5/11$
Expected return for A $= \text{Probability of A winning} \times \text{Amount of money on offer} \\=\frac{6}{11} \times 11 = 6.$
Expected return for B $= \text{Probability of B winning} \times \text{Amount of money on offer} \\=\frac{5}{11} \times 11 = 5.$
So, ans B)