Logistic regression is linear when the parameter, $\pi$, that controls the behavior of the Bernoulli response is transformed into a log odds:
$$
\ln\left(\frac{\pi_i}{1-\pi_i}\right) = \beta_0 + \beta_1x_i
$$
Your variable PASS is a vector of predicted probabilities. These can be converted into log odds using the LHS of the equation above. Once there, these should form a straight line as a function of RIT. Here is some R code to do this:
oPASS = PASS / (1-PASS)
loPASS = log(oPASS)
A plot of these values shows that there was some rounding in the predicted probabilities that you were given:
You can also see the issue if you look at the loPASS variable:
> loPASS
[1] -Inf -Inf -Inf -4.5951199 -4.5951199 -4.5951199
[7] -3.8918203 -3.1780538 -2.7515353 -2.1972246 -1.7346011 -1.2083112
[13] -0.7081851 -0.2006707 0.2818512 0.8001193 1.3249254 1.8152900
[19] 2.3136349 2.7515353 3.1780538 3.8918203 4.5951199 4.5951199
[25] Inf Inf Inf Inf Inf Inf
[31] Inf Inf Inf Inf Inf Inf
[37] Inf
Thus, we will work with the 7th & 23rd data points to get a reasonably accurate result.
Once we have these values, we can calculate the slope using the point-slope formula, and the intercept, by algebraically rearranging the equation of the line:
b1 = (loPASS[23]-loPASS[7]) / (RIT[23]-RIT[7])
b0 = loPASS[7] - b1*RIT[7]
That yields the parameter estimates that had been used to generate the predicted probabilities that you were given:
> b0
[1] -19.80483
> b1
[1] 0.1060868
For more information about logistic regression, it may help you to read my answer here: difference-between-logit-and-probit-models.