For this problem:

${\mathrm{D}}^3=\frac{{\displaystyle \frac{16}{\mathrm{\pi }}\frac{\mathrm{T}}{\mathrm{\tau }}}}{\left[1-{\left({\displaystyle \frac{\mathrm{d}}{\mathrm{D}}}\right)}^4\right]}=\frac{{\displaystyle \frac{16}{\mathrm{\pi }}\frac{(424.413 \mathrm{N}-\mathrm{m})(1,000 \mathrm{mm/m})}{75 \mathrm{N/mm}{}^2}}}{1-{(0.80)}^4}$

and therefore, the minimum outside diameter required to keep the shear stresses at or below the $75 \mathrm{MPa}$ allowable stress is:

$\therefore \mathrm{D}=36.547 \mathrm{mm}=\mathbf{36}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{mm}$