Given that, $2^{6x}+2^{3x+2}-21 =0$
$ \Rightarrow \left ( {2^{3x}} \right)^2 +2^{3x} \cdot 2^2-21 =0 \quad \longrightarrow (1)$
Let $2^{3x}$ be $k.$
Now, from equation $(1),$ we get
$ \Rightarrow k^2 +4k-21 =0$
$ \Rightarrow k^2 +7k -3k-21 =0$
$ \Rightarrow k(k+7)-3(k+7)=0$
$ \Rightarrow (k+7)(k-3)=0$
$ \Rightarrow \boxed {k=-7,3}$
$ \Rightarrow k \neq {-7}$, because $2^{3x}$ not be $-7$ for any value of $x.$
So, $ \boxed {k =3}$
$ \Rightarrow 2^{3x}=3$
Taking $\log_{2}$on the both sides.
$ \Rightarrow\log_{2}2^{3x} =\log_{2}3$
$ \Rightarrow 3x \log_{2}2 = \log_{2}3$
$ \Rightarrow 3x = \log_{2}3 \quad (\because \log_{a}a = 1)$
$ \Rightarrow \boxed{ x = \frac {\log_{2}3}{3}}$
Correct Answer: C