$\sigma = \dfrac{P}{A}$ — read aloud: sigma equals P over A. $P$ is the axial force the member carries, in newtons; $A$ is the cross-sectional area that carries it, measured ACROSS the member rather than along it; and $\sigma$ is the normal stress, the intensity of that force inside the material, in pascals. Whichever of the three the question leaves blank is the one you solve for.

Now the identity that turns this course into mental arithmetic: $1\ \mathrm{MPa} = 1\ \mathrm{N}/\mathrm{mm}^2$. Newtons over square millimetres come out as megapascals directly — no powers of ten, no table. That is why engineers quote sections in mm², and it is also why a load in KILOnewtons has to spend its thousand before it goes anywhere near the socket.

Push the units through every rearrangement. If they refuse to cancel into the answer's units, the rearrangement is wrong — no appeal. Units that DO work out never prove you right; the check runs one way only, and it is still the cheapest check you will ever run.