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 $=p=\frac{3}{6}$ and failure is $q=1-p=\frac{1}{2}$.

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

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

$\Rightarrow {}^{n}C_7{\left(\frac{1}{2}\right)}^7{\left(\frac{1}{2}\right)}^{n-7}={}^{n}C_9{\left(\frac{1}{2}\right)}^9{\left(\frac{1}{2}\right)}^{n-9}$

$\Rightarrow {}^{n}C_7={}^{n}C_9$

$\Rightarrow {}^{n}C_{n-7}={}^{n}C_9$

$\Rightarrow n = 16$

$\Rightarrow P(2 \mathrm{times} \mathrm{even} )={}^{16}C_2{\left(\frac{1}{2}\right)}^{14}{\left(\frac{1}{2}\right)}^2$

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

$\Rightarrow k=60$

Hence this is the required option.