For uniform shear stress τ on the annulus,$T=\int\tau r\,dA=2\pi\tau(R_o^3-R_i^3)/3$. Hence $\tau=3(50)/[2(3.14)(.15^3-.12^3)]=14.50$ kPa, matching **14.49,A** within source rounding. The radii, not diameters, enter the cubic formula.
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