Given

Strength of $HNO_3 = 70\%$

Volume of solution = 100mL

Volume of $HNO_3$ = 70mL

Specific gravity of $HNO_3 = 1.42gmL^{-1}$

Since $70\times1.42g$ of pure acid is present in 100mL

$\therefore$ 63g of pure acid present in = $\large\frac{100\times63}{70\times1.42}$

=63.38 mL

Hence answer is (b)