Put pressure inside a thin-walled cylinder and the wall is pulled two ways at once. The bigger of the two runs around the girth, and it is called the hoop or circumferential stress: $\sigma_{h} = \dfrac{pd}{2t}$. Read aloud: sigma-h equals p d over two t. $\sigma_{h}$ is the hoop stress in MPa; $p$ is the internal gauge pressure in MPa — gauge, so atmospheric pressure outside is already netted off; $d$ is the internal diameter in mm; and $t$ is the wall thickness in mm. Solve it for whichever one the question withholds.

You will meet the same relation written $\sigma_{h} = \dfrac{pr}{t}$, with $r$ the internal RADIUS. Those are the same equation — the 2 in the denominator is exactly the conversion from diameter to radius. What is not the same is feeding a diameter into the pr/t form, which doubles your answer and is the classic wrong number on this topic. Pick one form and read the given carefully.

One quiet gift: with p in MPa and both lengths in mm, the millimetres cancel and the answer arrives in MPa with no conversion owed. That is not luck — a megapascal IS a newton per square millimetre. Thin-wall, incidentally, means $d/t$ above about 20; below that the stress varies through the wall and this formula starts telling comfortable lies.