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By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
∴ 1.86x10^-4/5 = X / 3
∴X = 1.1 x 10^-4 m/s
∴the average rate of formation of Br2 = 1.1x10^-4 m/s
Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:

When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:

1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid