Take $E= 8x10^4$ MPa. Compare Euler critical load by taking $\sigma_c$= 550 MPa and $\alpha^{'}$ = (1/1600). For what length of the column the critical load by the Euler’s and Rankine formula will be equal to each other.

Since both ends are hinged, $L_c = L = 6\hspace{0.05cm}m$

$A = \frac{\pi}{4}(D^2 - d^2) = 13744\hspace{0.05cm}mm^2 = 13.74\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^2\\ L = 6\hspace{0.05cm}m = 6000\hspace{0.05cm}mm\\ E = 8\hspace{0.05cm}\times\hspace{0.05cm}10^4\\ \alpha = \frac{1}{1600}$

$I_{xx} = I_{yy} = \frac{\pi}{64}(D^4 - d^4)\\ \hspace{0.5cm} = 53.68\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^4\\ K = \sqrt{\frac{I_{min}}{A}} = \sqrt{\frac{53.68\hspace{0.05cm}\times\hspace{0.05cm}10^3}{13744}}\\ \hspace{0.05cm} = 62.50$

$P_{Euler} = \frac{\pi^2}{L_c^2}EI\\ \hspace{0.25cm} = \frac{\pi^2}{(6\hspace{0.05cm}\times\hspace{0.05cm}10^3)^2}\hspace{0.05cm}\times\hspace{0.05cm}8\hspace{0.05cm}\times\hspace{0.05cm}10^4\hspace{0.05cm}\times\hspace{0.05cm}53.68\hspace{0.05cm}\times\hspace{0.05cm}10^6\\ \hspace{0.25cm} = 1177334.14\hspace{0.05cm}N = 1.17\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}KN$

$P_{Rankine} = \frac{\sigma_{cr}\hspace{0.05cm}\times\hspace{0.05cm}A}{1 + \alpha (\frac{L_c}{K})^2}\\ \hspace{0.25cm} = \frac{550\hspace{0.05cm}\times\hspace{0.05cm}13744}{1 + \frac{1}{1600}(\frac{6\hspace{0.05cm}\times\hspace{0.05cm}10^3}{62.50})^2}\\ \hspace{0.25cm} = 1.11\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}KN$

For the same length, $P_{Euler} = P_{Rankine}\\ \frac{\pi^2}{L_c^2}EI = \frac{\sigma_c . A}{1 + \alpha (\frac{L_c^2}{62\hspace{0.05cm}\times\hspace{0.05cm}50^2})}\\ \frac{4.239\hspace{0.05cm}\times\hspace{0.05cm}10^13}{L_c^2} = 7557000\\ L_c = 2368.41\hspace{0.05cm}mm$