The question is asking you to find the :delta:Go in your equation. You have everything you need, remember to put your temperature in Kelvins. Kb is 1.38 x 10-23 J/K, h is6.63 x 10-34, and R is 8.314 J/mol*K.
To solve for the free energy of activation, you will need some properties of logarithms. Taking the natural log (ln) of an exponential will get rid of e. I would move the pre-exponent (kbT/H) to the other side first and then take the ln of each side to start.
Hope that helps.