Yes, and the first choice's reasoning is the only correct one.
The MVT guarantees the existence of at least one [tex]c\in(0,3)[/tex] such that
[tex]f'(c)=\dfrac{f(3)-f(0)}{3-0}[/tex]
We have [tex]f'(x)=-5e^{-5x}[/tex], so that
[tex]-5e^{-5c}=\dfrac{e^{-15}-1}3\implies c=-\dfrac15\ln\dfrac{1-e^{-15}}{15}[/tex]