When considering the Sun as a solid sphere, the formula for the moment of inertia is given by:

$ I = \frac{2}{5} m R^2 $

Here, $ m $ is the mass and $ R $ is the radius of the Sun.

To find the new period of revolution if the Sun expands to twice its current radius, we apply the conservation of angular momentum. The principle states that angular momentum before and after the expansion must be equal:

$ l' \omega' = l \omega $

Substituting the expression for angular momentum $ (I \times \omega) $ for both initial and expanded states, we get:

$ \frac{2}{5} m (2R)^2 \times \frac{2\pi}{T'} = \frac{2}{5} m R^2 \times \frac{2\pi}{T} $

This simplifies to:

$ \Rightarrow 4mR^2 \times \frac{2\pi}{T'} = mR^2 \times \frac{2\pi}{T} $

Cancelling out common terms, we find:

$ \Rightarrow 4 \times \frac{1}{T'} = \frac{1}{T} $

Thus:

$ \Rightarrow T' = 4T = 4 \times 27 = 108 \text{ days} $

Therefore, the new period of revolution would be 108 days if the Sun were to expand to twice its present radius, assuming it remains a sphere of uniform density and there are no external influences.