It should be double

Let $p$ be the probability of tossing a head. Then the probability of tossing a tail is $1 - p$ . There are $\dbinom{3}{2} = 3$ ways of obtaining $2$ heads and $1$ tail in three tosses, each of which occurs with probability $p^{2}(1 - p)$ . Thus we have that

$3p^{2}(1 - p) = \dfrac{1}{6} \Longrightarrow p^{2}(1 - p) = \dfrac{1}{18}$ .

Now the probability of tossing $4$ heads and $2$ tails in $6$ tosses is then

$\dbinom{6}{4}p^{4}(1 - p)^{2} = 15(p^{2}(1 - p))^{2} = 15 \times \left(\dfrac{1}{18}\right)^{2} = \dfrac{15}{18^{2}} = \dfrac{5}{108}$ ,

which does not correspond to any of the numerical options provided, and so the answer is "none of these choices".