0.2 M solution of a weak acid HA is 1% ionized ${25}^{0}C$.  $K_a$ for the acid is equal to

$\mathrm{HA} + {\mathrm{H}}_2\mathrm{O} \rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+} + {\mathrm{A}}^{-}$

$\left[\mathrm{HA}\right] =\mathrm{C}-\mathrm{C\alpha }$

$\left[\mathrm{HA}\right] = 0.2 - \frac{0.2 \times 1}{100} = 0.198 \mathrm{M}$

$\left[{\mathrm{A}}^{-1}\right] =\left[{\mathrm{H}}^{+}\right]=\mathrm{C\alpha }=0.2\times {10}^{-2}= 0.002 \mathrm{M}$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{A}}^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[\mathrm{HA}\right]}=\frac{0.002\times 0.002}{0.198}$