If the PH of a 0.227 M fluoroacetic acid solution is 2.03 at equilibrium you calculate the Ks

The pH is given as 2.03.
$[\ce{H+}] = 10^{-\text{pH}}$
$[\ce{H+}] = 10^{-2.03}$
$[\ce{H+}] = 0.0093325\ \text{M}$

Let HA represent fluoroacetic acid. The dissociation reaction is:
$\ce{HA(aq) <=> H+(aq) + A-(aq)}$

From Step 1, we know that $x = [\ce{H+}]_{\text{eq}} = 0.0093325\ \text{M}$.

$[\ce{H+}]_{\text{eq}} = 0.0093325\ \text{M}$
$[\ce{A-}]_{\text{eq}} = x = 0.0093325\ \text{M}$
$[\ce{HA}]_{\text{eq}} = 0.227 - x = 0.227 - 0.0093325 = 0.2176675\ \text{M}$

The expression for $K_a$ is:
$K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$
Substitute the equilibrium concentrations:
$K_a = \frac{(0.0093325)(0.0093325)}{(0.2176675)}$
$K_a = \frac{0.0000871055}{0.2176675}$
$K_a = 0.00040017$

Rounding to two significant figures based on the pH (2.03 has two decimal places, implying two significant figures in the concentration derived from it, or the initial concentration 0.227 M has 3 sig figs, so let's use 3 sig figs for the final answer).

$K_a = 4.00 \times 10^{-4}$

The $K_a$ for fluoroacetic acid is:
$4.00 \times 10^{-4}$