(i) Reaction quotient, ${\mathrm{Q}}_{\mathrm{c}}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]}$

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.
The given reaction is:

${\mathrm{CH}}_3{\mathrm{COOH}}_{\left(\mathrm{I}\right) } + {\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{\left(\mathrm{I}\right) } \leftrightarrow {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5\left(\mathrm{I}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{I}\right)}$
$\mathrm{Initial} \mathrm{conc}. \frac{\mathrm{I}}{\mathrm{V}}\mathrm{M} \frac{0.18}{\mathrm{V}}\mathrm{M} 0 0$
$\mathrm{At} \mathrm{equilibrium} \frac{1-0.171}{\mathrm{V}} \frac{0.18-0.171}{\mathrm{V}} \frac{0.171}{\mathrm{V}}\mathrm{M} \frac{0.171}{\mathrm{V}}\mathrm{M}$

Therefore, the equilibrium constant for the given reaction is:

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]}$
$=\frac{{\displaystyle \frac{0.171}{\mathrm{V}}}\times {\displaystyle \frac{0.171}{\mathrm{V}}}}{{\displaystyle \frac{0.829}{\mathrm{V}}}\times {\displaystyle \frac{0.009}{\mathrm{V}}}}=3.919$
$=3.92 \left(\mathrm{approx}.\right)$

(iii) Let the volume of the reaction mixture be V.

${\mathrm{CH}}_3{\mathrm{COOH}}_{\left(\mathrm{I}\right) } + {\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{\left(\mathrm{I}\right) } \leftrightarrow {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5\left(\mathrm{I}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{I}\right)}$
$\mathrm{Initial} \mathrm{conc}. \frac{1.0}{\mathrm{V}}\mathrm{M} \frac{0.5}{\mathrm{V}}\mathrm{M} 0 0$
$\mathrm{After} \mathrm{some} \mathrm{time} \frac{10-0.214}{\mathrm{V}} \frac{0.5-0.214}{\mathrm{V}} \frac{0.214}{\mathrm{V}}\mathrm{M} \frac{0.214}{\mathrm{V}}\mathrm{M}$
$=\frac{1.786}{\mathrm{V}}\mathrm{M} =\frac{0.286}{\mathrm{V}}\mathrm{M}$

Therefore, the reaction quotient is,

${\mathrm{Q}}_{\mathrm{c}}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]}$
$=\frac{{\displaystyle \frac{0.214}{\mathrm{V}}}\times {\displaystyle \frac{0.214}{\mathrm{V}}}}{{\displaystyle \frac{0.786}{\mathrm{V}}}\times {\displaystyle \frac{0.286}{\mathrm{V}}}}=0.2037$
$=0204 \left(\mathrm{approx}.\right)$

Since, $Q_c<K_c$, equilibrium has not been reached.