Let a die be rolled n times. Let the probability of getting odd...

Given that $P$(getting odd $7$ times) $= P$(getting odd $9$ times).

This is the case of binomial distribution with $n$ successes and probability of success $= p = \dfrac{3}{6}$ and failure is $q = 1 - p = \dfrac{1}{2}$.

We know that $P\left(X = k\right) = Ck n {\left(p\right)}^{k}{\left(q\right)}^{n - k}$

Now $P\left(X = 7\right) = P\left(X = 9\right)$

$⇒ C7 n {\left(\dfrac{1}{2}\right)}^{7}{\left(\dfrac{1}{2}\right)}^{n - 7} = C9 n {\left(\dfrac{1}{2}\right)}^{9}{\left(\dfrac{1}{2}\right)}^{n - 9}$

$⇒ C7 n = C9 n$

$⇒ Cn - 7 n = C9 n$

$⇒ n = 16$

$⇒ P ( 2timeseven ) = C2 16 {\left(\dfrac{1}{2}\right)}^{14}{\left(\dfrac{1}{2}\right)}^{2}$

$= \dfrac{15 \times 4}{{2}^{15}} = \dfrac{60}{{2}^{15}} = \dfrac{k}{{2}^{15}}$

$⇒ k = 60$

Hence this is the required option.