Calculate the torque transmitted by the shafts.

The power ($P$) transmitted is $60 \, \text{kW} = 60 \times 10^3 \, \text{W}$.
The rotational speed ($N$) is $240 \, \text{r/min}$.
First, convert the rotational speed to angular velocity ($\omega$) in rad/s:

$$N = 240 \, \text{r/min} = \frac{240}{60} \, \text{r/s} = 4 \, \text{r/s}$$ $$\omega = 2 \pi N = 2 \pi (4) \, \text{rad/s} = 8 \pi \, \text{rad/s}$$

Now, calculate the torque ($T$) using the formula $P = T \omega$:

$$T = \frac{P}{\omega} = \frac{60 \times 10^3 \, \text{W}}{8 \pi \, \text{rad/s}}$$ $$T = \frac{7500}{\pi} \, \text{N} \cdot \text{m}$$ $$T \approx 2387.32 \, \text{N} \cdot \text{m}$$

The claw coupling can be treated as a hollow shaft for torque transmission. The maximum shear stress ($\tau_{max}$) in a hollow circular shaft is given by:

$$\tau_{max} = \frac{T r_o}{J}$$

where $r_o$ is the outside radius and $J$ is the polar moment of inertia for a hollow shaft.
The outside radius is $r_o = d_o/2$.
The polar moment of inertia for a hollow shaft is $J = \frac{\pi}{32} (d_o^4 - d_i^4)$, where $d_o$ is the outside diameter and $d_i$ is the inside diameter.
Substitute these into the shear stress formula:

$$\tau_{max} = \frac{T (d_o/2)}{\frac{\pi}{32} (d_o^4 - d_i^4)} = \frac{16 T d_o}{\pi (d_o^4 - d_i^4)}$$

We are given that the outside diameter of the claw is twice the inside diameter, so $d_o = 2 d_i$.
The maximum allowable shear stress is $\tau_{max} = 6.4 \, \text{MPa} = 6.4 \times 10^6 \, \text{Pa}$.
Substitute $d_o = 2 d_i$ into the shear stress equation:

$$\tau_{max} = \frac{16 T (2 d_i)}{\pi ((2 d_i)^4 - d_i^4)}$$ $$\tau_{max} = \frac{32 T d_i}{\pi (16 d_i^4 - d_i^4)}$$ $$\tau_{max} = \frac{32 T d_i}{\pi (15 d_i^4)}$$ $$\tau_{max} = \frac{32 T}{15 \pi d_i^3}$$

Now, solve for $d_i$:

$$d_i^3 = \frac{32 T}{15 \pi \tau_{max}}$$

Substitute the values for $T$ and $\tau_{max}$:

$$d_i^3 = \frac{32 \times \left(\frac{7500}{\pi}\right)}{15 \pi (6.4 \times 10^6)}$$ $$d_i^3 = \frac{32 \times 7500}{15 \pi^2 \times 6.4 \times 10^6}$$ $$d_i^3 = \frac{240000}{96 \pi^2 \times 10^6}$$ $$d_i^3 = \frac{2500}{\pi^2 \times 10^6}$$ $$d_i^3 \approx \frac{2500}{9.8696 \times 10^6} \approx 0.000253302 \, \text{m}^3$$ $$d_i = (0.000253302)^{1/3} \, \text{m}$$ $$d_i \approx 0.06326 \, \text{m}$$

Convert to millimeters:

$$d_i \approx 63.26 \, \text{mm}$$

Now, calculate the outside diameter $d_o$:

$$d_o = 2 d_i = 2 \times 63.26 \, \text{mm}$$ $$d_o \approx 126.52 \, \text{mm}$$

The inside diameter of the claw is $\boxed{\text{63.26 mm}}$.
The outside diameter of the claw is $\boxed{\text{126.52 mm}}$.