Using mean value theorem we know that,
$f'(c)=\frac{f(b)-f(a)}{b-a}$
Here, $f(x)=x(x+3)e^{-\frac{x}{2}}$
$\implies f'(x)=e^{-\frac{x}{2}}(2x+3)+(x^{2}+3x)e^{-\frac{x}{2}}*(-\frac{1}{2})$
$\implies f'(x)=e^{-\frac{x}{2}}*\frac{x+6-x^{2}}{2}$
$\implies f'(c)=e^{-\frac{c}{2}}*\frac{c+6-c^{2}}{2}$
Here, the range $[a,b]$ is given as $[-3,0].$
$f(a)=f(-3)=0$
$f(b)=f(0)=0$
So,
$f'(c)=\frac{f(b)-f(a)}{b-a}$
$\implies f'(c)=\frac{f(0)-f(-3)}{0-(-3)}$
$\implies e^{-\frac{c}{2}}*\frac{(c+6-c^{2})}{2}=0$
now, $e^{-\frac{x}{2}}\neq 0$
So, $\frac{c+6-c^{2}}{2}=0$
$\implies c^{2}-c-6=0$
$\implies c^{2}-(3-2)c-6=0$
$\implies (c-3)(c+2)=0$
$\implies c=3$ or $c=-2$
now value of $c$ has to be in the range $[-3,0]$ so $c=3$ is not possible..
So, the value of $c$ is -2.