69 mass per cent of nitric acid means that 69 g of $\mathrm{HNO}_3$ is present in 100 g of solution. ∴ Volume of solution = $\frac{\text{Mass}}{\text{Density}}$ = $\frac{100\,\mathrm{g}}{1.41\,\mathrm{g}\,\mathrm{mL}^{-1}}$ = 70.92 mL Moles of $\mathrm{HNO}_3$ = $\frac{69}{63}$ = 1.095 Molarity = $\frac{\text{Moles of }\mathrm{HNO}_3}{\text{Volume of soln. in }\mathrm{mL}}$ × 1000 = $\frac{1.095}{70.92}$ × 1000 = 15.44 M