$f(x) = \left\{ {\begin{array}{cccccccccccccccccccc}{\frac{{{2^{x + 2}} - 16}}{{{4^x} - 16}},}&{{\rm{ if }}x \ne 2}\\{k,}&{{\rm{ if }}x = 2}\end{array}} \right.$at $x = 2$}

We have $f(x) = \left\{ {\begin{array}{cccccccccccccccccccc}{\frac{{{2^{x + 2}} - 16}}{{{4^x} - 16}},}&{{\rm{ if }}x \ne 2}\\{k,}&{{\rm{ if }}x = 2}\end{array}} \right.$at $x = 2$

Since, $f(x)$ is continuous at $x = 2$.

therefore,${\rm{LHL}} = {\rm{RHL}} = f(2)$

At $x = 2,$ $\mathop {\lim }\limits_{x \to 2} \frac{{{2^x} \cdot {2^2} - {2^4}}}{{{4^x} - {4^2}}} = \mathop {\lim }\limits_{x \to 2} \frac{{4 \cdot \left( {{2^x} - 4} \right)}}{{{{\left( {{2^x}} \right)}^2} - {{(4)}^2}}}$

$= \mathop {\lim }\limits_{x \to 2} \frac{{4 \cdot \left( {{2^x} - 4} \right)}}{{\left( {{2^x} - 4} \right)\left( {{2^x} + 4} \right)}}$

$= \mathop {\lim }\limits_{x \to 2} \frac{4}{{{2^x} + 4}} = \frac{4}{8} = \frac{1}{2}$

But $f(2) = k$

Therefore the value of $k = \frac{1}{2}$