I'm guessing it's section two Question 3
3a. (i) - As you can see from the diagram, position Y is more closer (in length/distance) to the steel sample than Position X therefore Position Y will be transmitted in less time than Position X. Therefore you can easily identify that the time taken between the pulse being transmitted and received at Position X is 15microseconds
(ii) - You want to calculate the length of the steel sample i.e it's width at position X (i don't know if I'm making sense but just let me know if i'm confusing you). You are given the speed of ultrasound in steel which is 5200ms-1. And you have also concluded from the previous question that the time for Position X to be transmitted and received is 15microseconds. Therefore you can use d=vt to calculate the distance of the steel sample.
Note: the time is in microseconds so do d=5200 x 15x10-6. This equals to 0.078m.
Then to calculate the thickness you divide 0.078 by 2 to get 0.039m.
b. So from the graphs and diagrams you have concluded that Position X has a longer distance than Position Y. However position Z is in between them. So you would just draw a line that is between Position X and Position Y to get the mark.
c. (i) - Simply do f=1/T. i.e f= 1/4.0x10-6 (because 4.0 is in microseconds) and you should get the answer 2.5x10^5
(ii) - You would use v=fWavelength
wavelength=v/f
= 5200/2.5x10^5
= 0.021m
d - The speed of ultrasound in brass will be less than steel because in brass it takes longer for it to travel the same thickness.
Hope this helps