The distance between object and its two times magnified real image as produced by a convex lens is $$45 \text{ cm}$$. The focal length of the lens used is ________ cm.

We need to find the focal length of a convex lens that produces a 2 times magnified real image, with the distance between the object and image being 45 cm.

For a real, inverted image with magnification $$|m| = 2$$, we have $$m = -2$$. Since $$m = \frac{v}{u}$$, it follows that $$v = -2u$$.

Adopting the sign convention by letting the object distance be $$u = -d$$ (negative because the object is on the left), we get $$v = -2(-d) = 2d\,. $$

The distance between object and image is given by $$|v - u| = |2d - (-d)| = 3d = 45\text{ cm}\,, $$ which yields $$d = 15\text{ cm}\,. $$ Hence $$u = -15\text{ cm}$$ and $$v = 30\text{ cm}\,. $$

Substituting these into the lens formula $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{30} - \frac{1}{-15} = \frac{1}{30} + \frac{1}{15} = \frac{3}{30} = \frac{1}{10}$$ gives $$f = 10\text{ cm}\,. $$

The focal length of the lens is $$\boxed{10}$$ cm.