Slide 1 of reply 3 shows the equilibria with hydroxide. Water and alcohol are nearly equal while HBr greatly favors bromide formation. Slide 2 shows the equilibria of the reaction in question, but from the reverse side. That is all you need to answer the question. It shows bromide opening of an expoxide is an uphill reaction, and more uphill than sulfide or alkoxide. Nucleophile A will have the largest amount of ring opening. This is indicated by Slide 2.
Certainly, mercaptides react to open epoxides and it would be difficult to know where an equilibrium might lie. However, it can also be difficult to know whether the product are also influenced by the reaction conditions. For example, the intent of reacting a sulfide with an epoxide might be presumed to form its product and avoid its reverse reaction. If it's reaction were carried out in a protic solvent, protonation of the alkoxide intermediate will stop the reverse reaction. Since the reaction only requires a catalytic amount of base, the mercaptan is the most acidic component and either directly or indirectly will form more mercaptide. This is different than the equilibrium shown on Slide 2. It shows a reaction of an alkoxide to form an epoxide plus mercaptide, the reverse reaction. Simply by formation of the conjugate base of a stronger acid, this reaction should be favored, however, steric strain can increase the energy level of the epoxide and reduce the equilibrium.
The other slides tell you about factors in the reaction, such as relief of strain. Fluoride was not on the slides, but it should have predictable properties. Acidity is data. It should tell you HBr is a stronger acid than HF. The arguments about atom size are an arguments to correct for the erroneous prediction that HF should be a stronger acid from electronegativity theory.