Answer
$0.68C^{\circ}$
Work Step by Step
Half of the initial gravitational potential energy goes toward heating the boulder.
$$\frac{1}{2}mgh=mc_{marble}\Delta T$$
$$\Delta T=\frac{gh}{2 c_{marble}}=\frac{(9.80m/s^2)(120m)}{2 (860 J/(kg\cdot C^{\circ}))}=0.68C^{\circ}$$