Is the gravity on Kepler 39b higher than on the Sun? According to the formula for gravitational acceleration $a = GM/r^2$ where $G=6.674 \times 10^{-11}$, Jupiter with mass 1.89813 × 10^27 kg and radius 69,911,000 m gives a gravity around 25m/s^2. The Sun with mass 1.989 × 10^30 kg and radius 695,510,000m will have a gravity around 270m/s^2.

But using the data for Kepler 39B, mass 18 times that of Jupiter 3.416634 × 10^28, and radius about 5/4 of Jupiter's 87,220,000m, I get almost 300m/s^2, higher than the Sun's. 

Is this right? 

I did and redid it, and still concerned that I am doing something wrong.
You are probably overlooking why the radius is what it is, or equivalently why the planet is denser than the Sun.  

The Sun is undergoing fusion in its core, which releases energy, which causes it to expand (though it achieves an equilibrium state, as expansion decreases rate of fusion which leads to contraction, which leads to accelerated fusion which leads to expansion...).  Jupiter-like bodies, however, don't produce heat from fusion (except maybe some deuterium fusion very early in their lives); they slowly cool and contract over time.  So such a body can compress into a denser state, allowing it achieve higher surface gravity (give or take difficulties in defining "surface" for a gaseous body).

Eyeballing the calculation, you can see that the sun is 10 times the radius of Jupiter and 1000 times the mass, so it's surface gravity should be about $1000/10^2=10$ times greater than Jupiter's, which is what you find.  But Kepler 39b is 18 times more massive than Jupiter and only marginally larger. That's a big uptick in density. So that it's surface gravity would be about $18/(1.25)^2=11.52$ times stronger than Jupiter's (also what you find).  And $11.52&gt;10$, so Kepler 39b has higher surface gravity.