In problems of this sort, the first thing to do is write a balanced chemical equation.
In this case,
Hg + Br2 ---> HgBr2
Remember that chemical equations are expressed in terms of MOLES, not GRAMS, so this equation says that
one mole of Hg (metal) plus one mole of bromine liquid (the standard state of bromine is as Br2, a red brown liquid) yields one mole of mercuric bromide.
The next thing to do is to determine the number of moles of mercury and of bromine you have. This requires a double conversion, the first from mL to grams (by using the density), then from grams to moles (by using the atomic mass).
For mercury, 5.00 mL * 13.6 g/mL = 68.0 g. 68.0g/(200.59 g/mole) = 0.339 moles Hg
For bromine, an extra step is needed. You have 5.00 mL of bromine, but at standard temperature and pressure, bromine consists of the Br2 molecule. So you have
5.00 mL * 3.10 g/mL = 15.5 g of Br2. 15.5 g Br2 * (2 Br/1 Br2) = 31.0 g Br
Now, 31.0 g Br/(79.904 g/mole) = 0.388 moles Br.
From the chemical equation, in order to convert all the mercury to mercuric bromide, you would need
0.339 moles of Br2 = 0.678 moles Br. However, you only have 0.388 moles Br.
Therefore, you need to run the problem the other way. Given 0.388 moles Br, you can make
0.388 moles Br *(1 Br2/2 Br) = 0.194 moles Br2
For each mole of Br2, you need one mole of Hg, so you need 0.194 moles Hg to react all of the available bromine to yield 0.194 moles HgBr2.
The molecular weight of HgBr2 is 360.398 g/mole, so you would wind up with
0.194 moles * 360.398 g/mole = 69.9 grams of HgBr2