Determine reaction order and reaction rate
The rate equation is $$\ln[\ce{Fe^{3+}}] = \ln[\ce{Fe^{3+}}]_0 − kt$$ So, $a-x=[A]$. Don't subtract the  next value. Put it as it is. That's just for your understanding, however. Don't try to use the integrated rate form to plot the graph. $[\ce{Fe^{3+}}]=238 \text{ where } t=10 \text{ and } [\ce{Fe^{3+}}]=227 \text{ where } t=20$. Using the above mentioned equation: $$\ln[A] = \ln[A]_0 − kt$$ Calculate the value of $\mathrm{\ln[\ce{Fe^{3+}}]_0}$ by eliminating k. Thereafter, you can graph the above mentioned equation $\ln[\ce{Fe^{3+}}] \text{ vs time t}$ with $\ln[\ce{Fe^{3+}}]_0$ as the y-intercept. The slope will yield the value of the rate constant.

[...]that would be the rate of the reaction?

For some reason unbeknownst to me, a lot of people tend to confuse between rate and rate constant. The slope will yield the rate constant. Next,
$$\textrm{rate}=k[\textrm A]^{a} \quad \text{}$$ where $a$ is the order of the reaction with respect to reactant $\ce{A}$.
As a side-note, since this is a pseudo-first order reaction, the rate law equation, initially this: $$\text{Rate} = k^{'}[\ce{Fe^{3+}}][\ce{H2O}]  $$
now appears as this:$$\text{Rate}=k[\ce{Fe^{3+}}]$$
the variable $'k'$ in the second equation is the slope that you obtained from the graph. Here $k=k^{'}[\ce{H2O}]_o$
For more about pseudo-first order reactions you can check one of my previous answers.