Step 1: Expand the actual AM signal instead of using the sideband formula directly.
An AM wave is $e(t) = A_c(1+m\cos\omega_m t)\cos\omega_c t$. Multiplying out, \[ e(t) = A_c\cos\omega_c t + mA_c\cos\omega_m t\cos\omega_c t \]
Step 2: Use the product-to-sum identity on the second term.
\[ \cos\omega_m t\cos\omega_c t = \frac{1}{2}\left[\cos(\omega_c+\omega_m)t + \cos(\omega_c-\omega_m)t\right] \] so \[ e(t) = A_c\cos\omega_c t + \frac{mA_c}{2}\cos(\omega_c+\omega_m)t + \frac{mA_c}{2}\cos(\omega_c-\omega_m)t \]
Step 3: Read off the sideband amplitude directly from this expansion.
The two sideband terms each carry amplitude $\frac{mA_c}{2}$; the factor of one half falls out naturally from the trig identity, rather than being an extra rule applied after computing $mA_c$.
Step 4: Substitute the given numbers.
\[ \frac{mA_c}{2} = \frac{0.6 \times 10}{2} = 3\ \text{V} \]
Step 5: Why the other options are wrong.
$6$ V is $mA_c$ without the halving the expansion requires; $10$ V is just the carrier amplitude; $16.6$ V does not follow from these numbers at all.
Final Answer:
\[ \boxed{3\ \text{V}} \]