Answer
$$W=2 \mathrm{e}^{x} \mathrm{e}^{2 x}.$$
Work Step by Step
The Wronskian of \left\{1, e^{x}, e^{2 x}\right\} is given by
$$W=\left|\begin{array}{ccc}1 &\mathrm{e}^{x} &\mathrm{e}^{2 x}\\ 0& \mathrm{e}^{x} &2\mathrm{e}^{2 x}\\ 0& \mathrm{e}^{x} &4 \mathrm{e}^{2 x}\end{array}\right|=2 \mathrm{e}^{x} \mathrm{e}^{2 x}.$$