If the internal and external diameter of the shafts are 125mm and 200mm respectively. Take $G= 84 x 10^3$ Mpa.

Given

$N = 180\hspace{0.05cm}rpm\\ P = 1600\hspace{0.05cm}KW = 1600\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}W\\ d = 125\hspace{0.05cm}mm,\hspace{0.5cm}r = 62.5\hspace{0.05cm}mm\\ D = 200\hspace{0.05cm}mm\hspace{0.5cm}R = 100\hspace{0.05cm}mm$

Solution

$P = \frac{2\pi NT}{60}\\ 1600\hspace{0.05cm}\times\hspace{0.05cm}10^3 = \frac{2\pi 180T}{60}\\ T = 84880\hspace{0.05cm}Nm = 84880\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Nmm$

Shear stress at outer radius (R = 100 mm)

$\frac{T}{J} = \frac{\tau}{R}\\ J = \frac{\pi}{32}[D^4 - d^4].....Hollow\\ \hspace{0.25cm} = \frac{\pi}{32}[200^4 - 125^4] = 133.11\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^4\\ \tau_{100} = \frac{TR}{J}\\ \hspace{0.25cm} = \frac{84880\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}100}{133.11\hspace{0.05cm}\times\hspace{0.05cm}10^6}\\ \tau_{100} = 63.76\hspace{0.05cm}N/mm^2$

$\tau$ at inner radius

$\tau_{62.5} = \frac{T.r}{J}\\ \tau_{62.5} = 39.85\hspace{0.05cm}N/mm^2$