If the sun were of uniform density throughout then the time of fall through the sun (or of any body of uniform density) is given by Pi/ sqrt( 4G rho Pi/3). Note that the time depends only on the density, not the size of the body. (Density is the relative mass of something per unit volume.) For the earth whose average denisty is about 5g/cm^3 the time is about 45 min (half the time of a near earth orbit of satellite.) The sun's density is is about 1 g/cm^3 so the time would be longer by about the square root of 5, or about 100 min.
However, the sun is not of uniform density. It is much less dense in its outer regions than inside. This would decrease the time. If all the mass were concentrated in the center, the time of free fall through the sun would go to 1/sqrt(2) of the above value (71 min). The actual value would depend on the exact density profile through the sun, which I do not happen to know off hand but it is probably closer to the latter (71min) than the former (100min). Of course this would also depend on where you defined the surface to be. The Sun is gaseous so there is no definite surface, the density just keeps falling. However the surface is usually defined as the optical surface--the radius at which the sun's light intensity suddenly drops drastically, which is what I used in the above.