Cubic Masterpiece

Let the smaller cubes have side length $1\,\mathrm{unit}$. So the original cube had side of length $3\,\mathrm{units}$ and as a cube has six faces it had a surface area of
$$6\times(3\,\mathrm{units}\times 3\,\mathrm{units})=54\,\mathrm{units}^2\;,$$
all of which was painted blue.

The total surface are of the 27 small cubes is
$$27\times 6\,\mathrm{units}^2 = 162\,\mathrm{units}^2\;.$$
So the required fraction is
$$\frac{54\,\mathrm{units}^2}{162\,\mathrm{units}^2} = \frac{1}{3}\;.$$