Answer
$4+\frac{x-16}{8}\Sigma_{n=2}^{\infty}(-1)^{n+1}\frac{1.3.5....(2n-3)(x-16)^{n}}{2^{5n-2}n!}$
and
$R=16$
Work Step by Step
$\sqrt x=4+\frac{x-16}{8}\Sigma_{n=2}^{\infty}(-1)^{n+1}\frac{1.3.5....(2n-3)(x-16)^{n}}{2^{5n-2}n!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{1.3.5....(2n-3)(2n-1)(x-16)^{n+1}}{2^{5(n+1)-2}(n+1)!}}{\frac{1.3.5....(2n-3)(x-16)^{n}}{2^{5n-2}n!}}|$
$=|\frac{x}{16}-1|\lt 1$
$-1\lt \frac{x}{16}-1 \lt 1$
$0 \lt x\lt 32$
The radius of convergence is always half of the width of the interval, so $R=16$.
