First calculate the number of moles of bromine and benzene.
Number of moles = 
Mass of benzene =
(given)
Molar mass of benzene=
Substitute the above values in formula, we get
Number of moles of benzene= 
= 
Mass of bromine = 69.3 g
Molar mass of bromine =
= 
Substitute the above values in formula, we get
Number of moles of benzene=
= 
Now, ratio of benzene and bromine comes out be 1:1 respectively and limiting reagent is benzene because present less as compared to bromine.
Thus, number of moles of bromobenzene = 
Theoretical yield = 
=
(molar mass of bromonenzene = 157.02 g/mol)
= 
Hence, theoretical yield of bromobenzene is 