If
[tex]M(t) = e^{3(e^t-1)}[/tex]
then using the chain rule, we have
[tex]M'(t) = e^{3(e^t-1)} \times 3e^t = 3e^{3(e^t-1)+t}[/tex]
[tex]M''(t) = 3e^{3(e^t-1)+t} \times (3e^t+1)[/tex]
Then
[tex]M''(0) = 3e^{3(e^0-1)+0} \times (3e^0+1) = 3e^0 \times (3+1) = \boxed{12}[/tex]
