A Taylor series about $x=a$ is given by
$$f(x) = \sum_{n=0}^\infty \frac{ f^{(n)}(a)}{n!} (x-a)^n$$
where $f^{(n)}(a)$ is the $n$h derivative of $f$ evaluated at $x=a$. The $0th$ derivative is just the function itself. To actually find the series you can try to compute the first several terms given from the series above and see if you discover a pattern. I'll get you started:
$$f(x) = \sum_{n=0}^\infty \frac{ f^{(n)}(16)}{n!} (x-16)^n$$
$$f(x) = \frac{f(16)}{0!} + \frac{f'(16)}{1!}(x-16) + \frac{f''(16)}{2!}(x-16)^2 + \cdots$$
Compute the derivatives above and see what you discover. For the radius of convergence, Mattos had a good suggestion in the comments.