Step 1: Combine the steps into one formula for pin diameter.
Starting from torque $T = \dfrac{P}{\omega}$, overload torque $T_{fail} = 1.5T$, force $F = \dfrac{T_{fail}}{r}$, and shear stress $\tau = \dfrac{F}{\pi d^2/4}$, these combine into $d = \sqrt{\dfrac{4 \times 1.5 \times P}{\pi \, r \, \tau \, \omega}}$.
Step 2: Work out $\omega$ first, since it appears inside the formula.
$\omega = \dfrac{2\pi N}{60} = \dfrac{2 \times 3.14 \times 540}{60} = 56.52$ rad/s.
Step 3: Substitute every value in one pass.
$d = \sqrt{\dfrac{4 \times 1.5 \times 40000}{3.14 \times 0.06 \times (200\times10^6) \times 56.52}}$.
Numerator $= 240000$. Denominator $= 3.14 \times 0.06 \times 200000000 \times 56.52 = 2.130\times10^{9}$.
$d = \sqrt{\dfrac{240000}{2.130\times10^{9}}} = \sqrt{1.127\times10^{-4}} = 0.01062$ m.
Step 4: Convert and confirm.
$d \approx 10.6$ mm, the same figure the step by step build up gives.
Final Answer:
The pin diameter nearest to the computed value is 10.6 mm, so option B is correct.
\[ \boxed{d \approx 10.6\ \text{mm}} \]