Your data looks something like this
t1 t2
12 8
7 10
... ...
So you have two variables with the two values for each object. What you need to do in case you want to use a dependent Wilcoxon test is to calculate the difference between the pairs and save the values in a new variable:
t1 t2 difference
12 8 4
7 10 -3
... ... ...
All you need to do is to calculate the median of the new variable difference. This gives you the value you are looking for. Also see here.
In R this would be
set.seed(1) # making it a reproducible example
n <- 100 # number of observations
t1 <- rnorm(n) # first measurement
t2 <- rnorm(n) # second measurement
difference <- t1 - t2 # difference between measurements
median(difference) # median of the difference
0.1955015 # this is the value you are looking for
If you need to run an independent Wilcoxon test, get the both medians of t1 and t2 first and then calculate the difference in the median. That would be:
# Difference in median between two groups
median(t1) - median(t2)
EDIT
@LuckyPal: As LuckyPal says the Wilcoxon test does not test the medians. This is most obvious if the median is near to zero but the Wilcoxon test is still significant (see here). Anyway, in my area it is nonetheless usual that people report the median to give some impression what the W statistic and the p value mean (how big is the change in the scale of the two variables?). This is why I tried to provide the answer the questioner was seemingly looking for.