HI

Start by drawing the diagram , which i think you have done so and this is what you should have in your diagram .

There are three forces here , one acting downwards through the centre of mass (It's right in the middle since it says weight of the rod is uniformly distributed ) , the other one is the reaction of his shoulder on the rod which acts upwards , and the last one would be the force he exerts downwards to keep the rod in equilibrium .

Lets call the centre , G , the point where the rod touches his shoulder , P .

(a) By taking moments about P , 3(9.81)(0.5)=F(0.3)

Calculate F

(b) Taking moments about G , R(0.5)=F(0.3+0.5)

Here , use the F you calculated above

OR

force going upwards = force going downwards .

R=3(9.81)+F