right i got the correct answer anyways-im sure im correct. the cross sectional area if it was fully hollow as in we wouldnt account for thickness t would be piR^2.naturally the question states that the engineers are testing hollow metal columns. now the thickness is the decrease in the HOLLOW cross sectional area as in a smaller cocentric circle- (which is a circle which originates at the same centre but of smaller or larger radius), with radius of (r-t) occurs where its area would be pi(r^2-2rt+t^2) . now we are assuming that R is much greater than t so t on its own is negligible in any calculation, it is too small on its own to make any noticeable effect so we can say t^2=0. hence subtracting the original cross sectional area from the smaller adjusted area for thickness we get piR^2-piR^2-0--2Rt= 2piRt. now we know pressure=force/area of cross section in contact which means we obtain the following. force due to the rectangular weight is mg. ASSUMPTION: only pressure is due to the rectangular mass and the hollow tube is said to be massless or very negligible mass, say aluminium. only assuming this because the question hadnt stated otherwise. thus pressure=mg/2piRt as required