Correct option is (a) 3.476 g
Normality of salt solution = \(\frac{3.92}{392} \times \frac 1{100} \times 1000 = 0.1 N\)
20 ml of 0.1 N salt solution \(\equiv\) 18 ml of KMnO4 solution
\(\therefore\) Normality of KMnO4 solution = \(\frac {20 \times 0.1}{18} = \frac 19 N\)
\(\therefore\) Strength of KMnO4 solution = \(\frac 19 \times 31.6 = 3.5 g L^{-1}\)