To determine the time required to consume half of A, we must first determine the rate law for the reaction based on the provided kinetic data.

From the data, we can write the general rate law for the reaction as:

$$ \text{Rate} = k [A]^m [B]^n $$

We can now use the experimental data to find the values of the exponents $ m $ and $ n $, as well as the rate constant $ k $.

From Experiment I and II, we can see that the initial concentration of A remains constant at 0.10 M while the concentration of B changes.

$$ \begin{align*} \text{Experiment I:} \quad \text{Rate} = k [0.10]^m [0.20]^n &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \\ \text{Experiment II:} \quad \text{Rate} = k [0.10]^m [0.25]^n &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \end{align*} $$

By comparing the rates from Experiment I and II:

$$ \frac{[0.10]^m [0.25]^n}{[0.10]^m [0.20]^n} = \frac{6.93 \times 10^{-3}}{6.93 \times 10^{-3}} $$

$$ \frac{[0.25]^n}{[0.20]^n} = 1 \implies \left(\frac{0.25}{0.20}\right)^n = 1 \implies n = 0 $$

Thus, the reaction is zero order with respect to B.

Next, we compare Experiment I and III to determine the order with respect to A:

$$ \begin{align*} \text{Experiment I:} \quad \text{Rate} = k [0.10]^m [0.20]^0 &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \\ \text{Experiment III:} \quad \text{Rate} = k [0.20]^m [0.30]^0 &= 1.386 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1} \end{align*} $$

$$ \frac{k [0.20]^m}{k [0.10]^m} = \frac{1.386 \times 10^{-2}}{6.93 \times 10^{-3}} $$

$$ \left( \frac{0.20}{0.10} \right)^m = 2 \implies (2)^m = 2 \implies m = 1 $$

Therefore, the reaction is first-order with respect to A.

Now, we can write the rate law as:

$$ \text{Rate} = k [A]^1 $$

Let's calculate the rate constant $ k $ using data from any experiment (say Experiment I):

$$ 6.93 \times 10^{-3} = k \times 0.10 $$

$$ k = \frac{6.93 \times 10^{-3}}{0.10} = 6.93 \times 10^{-2} \, \text{min}^{-1} $$

To determine the half-life (time required to consume half of A) for a first-order reaction, we use the formula:

$$ t_{1/2} = \frac{0.693}{k} $$

Substituting the value of $ k $:

$$ t_{1/2} = \frac{0.693}{6.93 \times 10^{-2}} = 10 \, \text{minutes} $$

Therefore, the time required to consume half of A is:

Option B: 10