1. Find the constant term condition

We need the constant term in $\left(a x^{1/8} + b x^{-1/12}\right)^{10}.$

The general term is $T_{r+1} = \binom{10}{r}(a x^{1/8})^{10-r}(b x^{-1/12})^r.$ So, $T_{r+1} = \binom{10}{r} a^{10-r} b^r x^{\frac{10-r}{8} - \frac{r}{12}}.$

For the constant term, exponent of $x$ must be $0$: $\frac{10-r}{8} - \frac{r}{12} = 0.$ Taking LCM $24$, $3(10-r) - 2r = 0$ $30 - 3r - 2r = 0$ $30 - 5r = 0$ $r = 6.$

Hence the constant term is $\binom{10}{6} a^{4} b^{6} = 210a^4b^6.$

So we must minimize $210 a^4 b^6$ given $\frac{1}{a^2} + \frac{1}{b^3} = 4.$

2. Substitute simpler variables

Let $x = \frac{1}{a^2}, \qquad y = \frac{1}{b^3}.$ Then $x,y>0$ and $x+y=4.$ Also, $a^4 = \frac{1}{x^2}, \qquad b^6 = \frac{1}{y^2}.$ Therefore $a^4b^6 = \frac{1}{x^2y^2}.$

So we need to minimize $210\cdot \frac{1}{x^2y^2},$ subject to $x+y=4$.

This is equivalent to maximizing $xy$.

3. Maximize $xy$ using AM-GM

For positive $x,y$ with fixed sum $x+y=4$, $xy \le \left(\frac{x+y}{2}\right)^2 = 2^2 = 4,$ with equality when $x=y=2.$

Hence $\frac{1}{x^2y^2} \ge \frac{1}{4^2} = \frac{1}{16}.$ So minimum of the constant term is $210\cdot \frac{1}{16} = \frac{105}{8}.$

4. Check equality case

Equality occurs when $\frac{1}{a^2} = \frac{1}{b^3} = 2,$ which is possible for positive real $a,b$. Hence the minimum is attained.

Therefore the minimum value of the constant term is $\boxed{\frac{105}{8}}.$

5. Compare with stored answer

Stored correct answer: C

Option C is $\frac{105}{8},$ which matches our result.