Quantity of charge passed = 2 $$\times$$ 10^$$-$$3 $$\times$$ 16 $$\times$$ 60 = 1.92 C

1.92 C charge $$ \equiv {1 \over {96500}} \times 1.92 \equiv 1.99 \times {10^{ - 5}}$$ mol of electrons

In the reaction, $$C{u^{2 + }}(aq) + 2e \to Cu(s)$$

2 mol of electrons = 1 mol of Cu²⁺ ions discharged

$$\therefore$$ 1.99 $$\times$$ 10^$$-$$5 mol of electrons = 9.95 $$\times$$ 10^$$-$$6 of Cu²⁺ ions discharged.

Absorbance of a solution is directly proportional to the concentration of the solution. 50% decrease of absorbance of the solution means 50% of the concentration of the solution is reduced. Hence, initial number of mol of Cu²⁺ ions = 2 $$\times$$ 9.95 $$\times$$ 10^$$-$$6 = 1.99 $$\times$$ 10^$$-$$5 mol.

$$\therefore$$ Initial concentration of Cu²⁺

i.e., $$CuS{O_4} = {{1.99 \times {{10}^{ - 5}}} \over {250}} \times 1000\,M = 7.96 \times {10^{ - 5}}\,M$$