Ok, I think I got the answer.
If you sketch a graph of $\displaystyle y= e^{2x}$ and that of $\displaystyle y = k\sqrt{x}$, you'll notice that if they meet at a point only, their gradients are the same.
This means, their derivative are equal.
$\displaystyle e^{2x} = k\sqrt{x}$
$\displaystyle 2e^{2x} = \dfrac{k}{2\sqrt{x}}$
$\displaystyle k = 4\sqrt{x}e^{2x}$
Plugging that in our equation:
$\displaystyle e^{2x} = (4\sqrt{x}e^{2x})\sqrt{x}$
$\displaystyle 1 = 4x$
$\displaystyle x = \dfrac14$
Now, plugging this value of x into the first equation:
$\displaystyle e^{2\frac14} = k\sqrt{\frac14}$
$\displaystyle e^{\frac12} = \frac{k}{2}$
$\displaystyle k = 2\sqrt{e}$