Given, Strength of $HNO_3 = 70\%$
Volume of the solution = 100mL
Volume of $HNO_3$ = 70 mL
$\therefore$ Weight of $HNO_3$ in the solution = $70\times1.42g$
$\therefore$ Equivalent weight of $HNO_3$ in the solution = $\large\frac{70\times1.42}{63}$
Since N=$\large\frac{\text{Equivalent weight of $HNO_3$ in the solution}}{\text{V in litre}}$
$N_{HNO_3} = \large\frac{70\times1.42}{63\times\large\frac{100}{1000}} = 15.78$
Hence answer is (a)