1. Define Ratios:
Let
$$\frac{A'C}{BA'} = \alpha, \quad \frac{B'A}{CB'} = \beta, \quad \frac{C'B}{AC'} = \gamma.$$
According to Ceva's Theorem, for the cevians $AA'$, $BB'$, and $CC'$ to be concurrent, the product of these ratios must be equal to 1:
$$\alpha \beta \gamma = 1.$$

2. Apply Van Aubel's Theorem:
Van Aubel's Theorem states that:
$$\frac{AO}{OA'} = \beta + \frac{1}{\gamma}, \quad \frac{BO}{OB'} = \gamma + \frac{1}{\alpha}, \quad \frac{CO}{OC'} = \alpha + \frac{1}{\beta}.$$

3. Sum of Ratios:
Given in the problem:
$$\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'} = 92.$$

4. Product of Ratios:
We need to find the value of:
$$\frac{AO}{OA'} \times \frac{BO}{OB'} \times \frac{CO}{OC'}.$$
Using the expressions from Van Aubel's Theorem, we have:
$$\frac{AO}{OA'} \times \frac{BO}{OB'} \times \frac{CO}{OC'} = \left(\beta + \frac{1}{\gamma}\right) \left(\gamma + \frac{1}{\alpha}\right) \left(\alpha + \frac{1}{\beta}\right).$$

5. Expand the Product:
Expanding the product, we get:
$$\left(\beta + \frac{1}{\gamma}\right) \left(\gamma + \frac{1}{\alpha}\right) \left(\alpha + \frac{1}{\beta}\right).$$
Using the identity $\alpha \beta \gamma = 1$, we can simplify:
$$\left(\beta + \frac{1}{\gamma}\right) \left(\gamma + \frac{1}{\alpha}\right) \left(\alpha + \frac{1}{\beta}\right) = \beta \gamma \alpha + \beta \gamma \frac{1}{\beta} + \beta \frac{1}{\alpha} \alpha + \frac{1}{\gamma} \gamma \alpha + \frac{1}{\gamma} \gamma \frac{1}{\alpha} + \frac{1}{\gamma} \frac{1}{\beta} \alpha + \frac{1}{\gamma} \frac{1}{\beta} \frac{1}{\alpha}.$$
Simplifying further:
$$= \alpha \beta \gamma + \gamma + \beta + \alpha + \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\alpha \beta \gamma}.$$
Since $\alpha \beta \gamma = 1$, we have:
$$= 1 + \gamma + \beta + \alpha + \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + 1.$$
Combining terms:
$$= 2 + \left(\beta + \frac{1}{\gamma}\right) + \left(\gamma + \frac{1}{\alpha}\right) + \left(\alpha + \frac{1}{\beta}\right).$$
Given that:
$$\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'} = 92,$$
we substitute:
$$= 2 + 92 = 94.$$

The final answer is $\boxed{94}$.