The general expansion of a function using taylor series at any given point $x=a$ for a function $f(x)$ is given as :
$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$ (Refer: Taylor Series Definition)
So the coefficient of the term $(x-3)^{5}$ would be $\frac{f'''''(3)}{5!}$
Here, $f(x)=e^{\frac{x}{2}}$ $\Rightarrow$ $f'(x)=e^{\frac{x}{2}}\times (\frac{1}{2})$
$\Rightarrow$ $f''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{2}$
$\Rightarrow$ $f'''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{3}$
So, similarly $f'''''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{5}$
$\therefore$ at $x=3$, the value of the coefficient would be : $e^{\frac{3}{2}}\times \frac{1}{2^{5}5!}$.
Option B is the correct answer.