$\theta_{\text{of 6.5' shaft}} + \theta_{\text{of 3.25' shaft}} = 6^\circ$
$\left( \dfrac{TL}{JG} \right)_{\text{of 6.5' shaft}} + \left( \dfrac{TL}{JG} \right)_{\text{of 3.25' shaft}} = 6^\circ \left( \dfrac{\pi}{180^\circ} \right)$
$\dfrac{T(6.5)(12^2)}{\frac{1}{32}\pi (2^4)(12 \times 10^6)} + \dfrac{T(3.25)(12^2)}{\frac{1}{32}\pi (1.5^4)(12 \times 10^6)} = \dfrac{\pi}{30}$
$T = 817.32 \, \text{lb}\cdot\text{ft}$
$\tau_{max} = \dfrac{16T}{\pi D^3}$
$\tau_{\text{of 6.5' shaft}} = \dfrac{16(817.32)(12)}{\pi (2^3)} = 6243.86 \, \text{ psi}$ answer
$\tau_{\text{of 3.25' shaft}} = \dfrac{16(817.32)(12)}{\pi (1.5^3)} = 14\,800.27 \, \text{ psi}$ answer