The volume of water needs to be added to 10.0 mL of nitric acid (density : $1.40\mathrm{g}{\mathrm{mL}}^{-1}$) ) containing 70 mass percent of acid to prepare 1.0 M solution would be

Mass of 10.0 mL of nitric acid, $m=V\rho =\left(10.0\mathrm{mL}\right)(1.40\mathrm{gmL}-1)=14.0\mathrm{g}$

Mass of nitric acid in this solution $=\left(14.0\mathrm{g}\right)(70/100)=9.8\mathrm{g}$

Amount of nitric acid, $n=m/M=9.8\mathrm{g}/63\mathrm{g}{\mathrm{mol}}^{-1}=0.1556\mathrm{mol}$

Molarity of solution $M=\frac{\mathrm{n}}{\mathrm{V}}=\left(\frac{0.1556\mathrm{mol}}{0.01\mathrm{L}}\right)=15.56\mathrm{M}$

This solution is to be diluted to prepare 1.0 M solution.

Volume of the resultant solution  $=\frac{\left(10\mathrm{mL}\right)\left(15.56\mathrm{M}\right)}{\left(1.0\mathrm{M}\right)}=155.6\mathrm{mL}$

Volume of water to be added $=155.6\mathrm{mL}-10.0\mathrm{mL}=145.6\mathrm{mL}$