Step-by-step Solution:

Correct Answer: A (64)

This problem is solved by simplifying the complex trigonometric product on the left-hand side (LHS) of the equation, solving for the value of $a$, and then calculating $3a^{-1}$.

Step 1: Simplify the Left-Hand Side (LHS)

The given expression is $LHS = \cos^2(10^\circ) \cos(20^\circ) \cos(40^\circ) \cos(50^\circ) \cos(70^\circ)$.

We can simplify this by grouping terms and using the identity $\cos(\theta)\cos(60^\circ - \theta)\cos(60^\circ + \theta) = \frac{1}{4}\cos(3\theta)$.

Let's rearrange the LHS:

$LHS = \cos(20^\circ)\cos(40^\circ) \cdot \cos(10^\circ)[\cos(10^\circ)\cos(50^\circ)\cos(70^\circ)]$

Notice that $\cos(50^\circ) = \cos(60^\circ - 10^\circ)$ and $\cos(70^\circ) = \cos(60^\circ + 10^\circ)$. Applying the identity with $\theta=10^\circ$ to the term in the brackets:

$[\cos(10^\circ)\cos(60^\circ - 10^\circ)\cos(60^\circ + 10^\circ)] = \frac{1}{4}\cos(3 \times 10^\circ) = \frac{1}{4}\cos(30^\circ) = \frac{1}{4}\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$

Substitute this back into the expression for the LHS:

$LHS = \frac{\sqrt{3}}{8} \cos(10^\circ)\cos(20^\circ)\cos(40^\circ)$

Step 2: Solve for the value of $a$

Now we set our simplified LHS equal to the given right-hand side (RHS):

$\frac{\sqrt{3}}{8} \cos(10^\circ)\cos(20^\circ)\cos(40^\circ) = a + \frac{\sqrt{3}}{16}\cos(10^\circ)$

Isolate $a$:

$a = \frac{\sqrt{3}}{8} \cos(10^\circ)\cos(20^\circ)\cos(40^\circ) - \frac{\sqrt{3}}{16}\cos(10^\circ)$

Factor out $\frac{\sqrt{3}}{16}\cos(10^\circ)$:

$a = \frac{\sqrt{3}}{16}\cos(10^\circ) [2\cos(20^\circ)\cos(40^\circ) - 1]$

Using the product-to-sum identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:

$2\cos(20^\circ)\cos(40^\circ) = \cos(60^\circ) + \cos(20^\circ) = \frac{1}{2} + \cos(20^\circ)$

Substitute this into the equation for $a$:

$a = \frac{\sqrt{3}}{16}\cos(10^\circ) \left[ \left(\frac{1}{2} + \cos(20^\circ)\right) - 1 \right] = \frac{\sqrt{3}}{16}\cos(10^\circ) \left[ \cos(20^\circ) - \frac{1}{2} \right]$

Using the double-angle identity $\cos(20^\circ) = 2\cos^2(10^\circ) - 1$:

$a = \frac{\sqrt{3}}{16}\cos(10^\circ) \left[ (2\cos^2(10^\circ) - 1) - \frac{1}{2} \right] = \frac{\sqrt{3}}{16}\cos(10^\circ) \left[ 2\cos^2(10^\circ) - \frac{3}{2} \right]$

$a = \frac{\sqrt{3}}{32} [4\cos^3(10^\circ) - 3\cos(10^\circ)]$

The term in the brackets is the triple-angle identity for cosine, $4\cos^3\theta - 3\cos\theta = \cos(3\theta)$.

$a = \frac{\sqrt{3}}{32} \cos(3 \times 10^\circ) = \frac{\sqrt{3}}{32} \cos(30^\circ)$

$a = \frac{\sqrt{3}}{32} \cdot \frac{\sqrt{3}}{2} = \frac{3}{64}$

Step 3: Calculate the Final Value

We need to find the value of $3a^{-1}$.

$a^{-1} = \frac{1}{a} = \frac{64}{3}$

$3a^{-1} = 3 \times \frac{64}{3} = 64$