The concentration of acetic acid required to get $3.5\times {10}^{-4}$ mole/lit of ${\mathrm{H}}^{+}$ion is$\left[{\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5}\right]$

 $\left[{\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5}\right]$,$3.5\times {10}^{-4}$ mole/lit of ${\mathrm{H}}^{+}$ion

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

 ${\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}$

 ${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

Since 1 mole of ${\mathrm{CH}}_3\mathrm{COOH}$ gives 1 mole of ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$and 1 mole of ${\mathrm{H}}^{+}$ions.

$\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]=\left[{\mathrm{H}}^{+}\right]$

${\mathrm{K}}_{\mathrm{a}}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

$\because {\mathrm{K}}_{\mathrm{a}}=1.8\times {10}^{-5},{\mathrm{H}}^{+}=3.5\times {10}^{-4} \mathrm{mol}/\mathrm{L}$

$1.8\times {10}^{-5}=\frac{{\left[3.5\times {10}^{-4}\right]}^2}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

$\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=\frac{12.25\times {10}^{-8}}{1.8\times {10}^{-5}}$

$=6.8\times {10}^{-3} \mathrm{mol}/\mathrm{L}$

Therefore, the correct option is (A).