Clausius-Clapeyron - Slope of the best-fit-line
Using your formula, $\mathrm{slope}=\frac{y_2-y_1}{x_2-x_1}$, and eyeballing the most extreme points on the chart, $\mathrm{slope} = \frac{0-6.2}{0.005 - 0.0033}$, I get a slope of 3647 K, pretty close to the "official" answer. 

If you want to get more fancy, you could use a software program to fit this data.

my_data &lt;- data.frame(vapor_pressure = c(1.1, 6, 28, 95, 293, 521),
                      temperature = c(200, 220, 240, 260, 280, 300))

# linear regression on transformed variables
my_data$ln_vp &lt;- log(my_data$vapor_pressure)
my_data$inverse_T &lt;- 1/(my_data$temperature)
linear_fit &lt;- lm(data = my_data, formula = ln_vp ~ inverse_T)
linear_fit$coefficients['inverse_T']




inverse_T: -3787.46829114867


This answer is even closer to the "official" answer.  The remaining difference is probably due to rounding errors or differences in fitting algorithms used.  It only affects the answer by 0.3%.