Identify given parameters and relationships.

The given parameters are:

• Inner diameter of bronze sleeve (and outer diameter of steel shaft), $d_i = d_s = 80 \, \text{mm} = 0.08 \, \text{m}$.
• Torque carried by bronze sleeve $T_b = 2 T_s$.
• Shear modulus of steel $G_s = 2.5 G_b$.
• Maximum allowable shear stress in steel $\tau_{s, max} = 60 \, \text{MPa} = 60 \times 10^6 \, \text{Pa}$.
• Maximum allowable shear stress in bronze $\tau_{b, max} = 40 \, \text{MPa} = 40 \times 10^6 \, \text{Pa}$.
• Rotation speed $N = 200 \, \text{r/min}$.

For a compound shaft, the angle of twist per unit length ($\theta$) is the same for both components:

$$\frac{T_s}{G_s J_s} = \frac{T_b}{G_b J_b}$$

Substitute the given relationships $T_b = 2 T_s$ and $G_s = 2.5 G_b$:

$$\frac{T_s}{2.5 G_b J_s} = \frac{2 T_s}{G_b J_b}$$ $$\frac{1}{2.5 J_s} = \frac{2}{J_b}$$ $$J_b = 5 J_s$$

The polar moment of inertia for a solid shaft is $J = \frac{\pi}{32} d^4$. For a hollow shaft, $J = \frac{\pi}{32} (d_o^4 - d_i^4)$.
For the steel shaft (solid): $J_s = \frac{\pi}{32} d_s^4$.
For the bronze sleeve (hollow): $J_b = \frac{\pi}{32} (d_o^4 - d_i^4)$.
Substitute these into $J_b = 5 J_s$:

$$\frac{\pi}{32} (d_o^4 - d_i^4) = 5 \frac{\pi}{32} d_s^4$$ $$d_o^4 - d_i^4 = 5 d_s^4$$

Since $d_i = d_s$:

$$d_o^4 - d_s^4 = 5 d_s^4$$ $$d_o^4 = 6 d_s^4$$ $$d_o = (6)^{1/4} d_s$$

Substitute $d_s = 80 \, \text{mm}$:

$$d_o = (6)^{1/4} \times 80 \, \text{mm} \approx 1.56508 \times 80 \, \text{mm}$$ $$d_o \approx 125.206 \, \text{mm}$$

The outside diameter of the sleeve is $\boxed{\text{125.21 mm}}$.

The shear stress in a shaft is given by $\tau = \frac{Tr}{J}$. For the maximum stress, $r$ is the outermost radius.
For the steel shaft: $\tau_s = \frac{T_s (d_s/2)}{J_s} = \frac{T_s (d_s/2)}{\frac{\pi}{32} d_s^4} = \frac{16 T_s}{\pi d_s^3}$.
For the bronze sleeve: $\tau_b = \frac{T_b (d_o/2)}{J_b} = \frac{T_b (d_o/2)}{\frac{\pi}{32} (d_o^4 - d_i^4)} = \frac{16 T_b d_o}{\pi (d_o^4 - d_i^4)}$.
From Step 2, we know $T_b = 2 T_s$, $d_o = (6)^{1/4} d_s$, and $d_o^4 - d_i^4 = 5 d_s^4$.
Let's find the ratio of the stresses:

$$\frac{\tau_b}{\tau_s} = \frac{\frac{16 T_b d_o}{\pi (d_o^4 - d_i^4)}}{\frac{16 T_s}{\pi d_s^3}} = \frac{T_b d_o d_s^3}{T_s (d_o^4 - d_i^4)}$$

Substitute the relationships:

$$\frac{\tau_b}{\tau_s} = \frac{(2 T_s) ((6)^{1/4} d_s) d_s^3}{T_s (5 d_s^4)} = \frac{2 (6)^{1/4}}{5}$$ $$\frac{\tau_b}{\tau_s} \approx \frac{2 \times 1.56508}{5} \approx 0.626032$$

So, $\tau_b \approx 0.626032 \tau_s$.
Now, check which material limits the torque:

• If $\tau_s = \tau_{s, max} = 60 \, \text{MPa}$, then $\tau_b = 0.626032 \times 60 \, \text{MPa} \approx 37.56 \, \text{MPa}$. This is less than $\tau_{b, max} = 40 \, \text{MPa}$, so it is acceptable.
• If $\tau_b = \tau_{b, max} = 40 \, \text{MPa}$, then $\tau_s = \frac{40 \, \text{MPa}}{0.626032} \approx 63.89 \, \text{MPa}$. This is greater than $\tau_{s, max} = 60 \, \text{MPa}$, so it is not acceptable. Therefore, the maximum torque is limited by the steel shaft reaching its maximum allowable shear stress, $\tau_s = 60 \, \text{MPa}$.

Calculate the torque carried by the steel shaft ($T_s$):

$$T_s = \frac{\pi d_s^3 \tau_s}{16} = \frac{\pi (0.08 \, \text{m})^3 (60 \times 10^6 \, \text{Pa})}{16}$$ $$T_s = \frac{\pi \times 0.000512 \, \text{m}^3 \times 60 \times 10^6 \, \text{Pa}}{16} = 1920 \pi \, \text{N} \cdot \text{m}$$ $$T_s \approx 6031.858 \, \text{N} \cdot \text{m}$$

Calculate the torque carried by the bronze sleeve ($T_b$):

$$T_b = 2 T_s = 2 \times 1920 \pi \, \text{N} \cdot \text{m} = 3840 \pi \, \text{N} \cdot \text{m}$$ $$T_b \approx 12063.716 \, \text{N} \cdot \text{m}$$

The total torque is the sum of the torques carried by each component:

$$T_{total} = T_s + T_b = 1920 \pi \, \text{N} \cdot \text{m} + 3840 \pi \, \text{N} \cdot \text{m} = 5760 \pi \, \text{N} \cdot \text{m}$$ $$T_{total} \approx 18095.574 \, \text{N} \cdot \text{m}$$

The torque that may be transmitted is $\boxed{\text{18.10 kN}\cdot\text{m}}$.

First, convert the rotation speed from r/min to rad/s:

$$N = 200 \, \text{r/min} = \frac{200}{60} \, \text{r/s} = \frac{10}{3} \, \text{r/s}$$

The angular velocity $\omega$ is:

$$\omega = 2 \pi N = 2 \pi \times \frac{10}{3} \, \text{rad/s} = \frac{20 \pi}{3} \, \text{rad/s}$$ $$\omega \approx 20.94395 \, \text{rad/s}$$

The power transmitted $P$ is given by $P = T_{total} \omega$:

$$P = (5760 \pi \, \text{N} \cdot \text{m}) \times \left(\frac{20 \pi}{3} \, \text{rad/s}\right)$$ $$P = \frac{5760 \times 20 \pi^2}{3} \, \text{W} = 1920 \times 20 \pi^2 \, \text{W} = 38400 \pi^2 \, \text{W}$$ $$P \approx 38400 \times (3.14159)^2 \, \text{W} \approx 379800.36 \, \text{W}$$ $$P \approx 379.80 \, \text{kW}$$

The power transmitted is $\boxed{\text{379.80 kW}}$.