Step 1: Understand the target.
We must find the value of $\sin20^\circ\sin40^\circ\sin80^\circ$. These three angles look special, so a standard formula should help.
Step 2: Recall the handy formula.
There is a known product rule: \[ \sin\theta\,\sin(60^\circ-\theta)\,\sin(60^\circ+\theta)=\frac{1}{4}\sin3\theta. \] This is the cleanest way to handle products spaced by $60^\circ$.
Step 3: Fit our angles to the formula.
Take $\theta=20^\circ$. Then $60^\circ-\theta=40^\circ$ and $60^\circ+\theta=80^\circ$. So our product is exactly the left side of the formula.
Step 4: Apply the formula.
So \[ \sin20^\circ\sin40^\circ\sin80^\circ=\frac{1}{4}\sin(3\times20^\circ)=\frac{1}{4}\sin60^\circ. \]
Step 5: Put in the known value.
We know $\sin60^\circ=\tfrac{\sqrt{3}}{2}$. So \[ \frac{1}{4}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}. \]
Step 6: State the result.
The product equals $\tfrac{\sqrt{3}}{8}$, which matches the marked option.
\[ \boxed{\dfrac{\sqrt{3}}{8}} \]