Calculate the equilibrium concentration of G6P from the first reaction.

The first reaction is:
$C_6H_{12}O_6(aq) + ATP(aq) \rightleftharpoons G6P(aq) + ADP(aq)$
The equilibrium constant for this reaction is $K_1 = 4.63 \times 10^3$.
The equilibrium expression is:

$$K_1 = \frac{[G6P][ADP]}{[C_6H_{12}O_6][ATP]}$$

We are given the following equilibrium concentrations:
$[C_6H_{12}O_6] = 1.2 \times 10^{-6} \text{ M}$
$[ATP] = 1.0 \times 10^{-4} \text{ M}$
$[ADP] = 1.0 \times 10^{-2} \text{ M}$
Rearrange the equation to solve for $[G6P]$:

$$[G6P] = K_1 \times \frac{[C_6H_{12}O_6][ATP]}{[ADP]}$$

Substitute the given values:

$$[G6P] = (4.63 \times 10^3) \times \frac{(1.2 \times 10^{-6} \text{ M})(1.0 \times 10^{-4} \text{ M})}{(1.0 \times 10^{-2} \text{ M})}$$ $$[G6P] = (4.63 \times 10^3) \times \frac{1.2 \times 10^{-10}}{1.0 \times 10^{-2}}$$ $$[G6P] = (4.63 \times 10^3) \times (1.2 \times 10^{-8})$$ $$[G6P] = 5.556 \times 10^{-5} \text{ M}$$

The second reaction is:
$G6P(aq) \rightleftharpoons F6P(aq)$
The equilibrium constant for this reaction is $K_2 = 2.76 \times 10^{-1}$.
The equilibrium expression is:

$$K_2 = \frac{[F6P]}{[G6P]}$$

We use the equilibrium concentration of $[G6P]$ calculated in Step 1. Rearrange the equation to solve for $[F6P]$:

$$[F6P] = K_2 \times [G6P]$$

Substitute the values:

$$[F6P] = (2.76 \times 10^{-1}) \times (5.556 \times 10^{-5} \text{ M})$$ $$[F6P] = 1.533456 \times 10^{-5} \text{ M}$$

The given concentrations have 2 significant figures ($1.2 \times 10^{-6}$ M, $1.0 \times 10^{-4}$ M, $1.0 \times 10^{-2}$ M). Therefore, the final answer should be rounded to 2 significant figures.

$$[F6P] \approx 1.5 \times 10^{-5} \text{ M}$$

The equilibrium concentration of F6P is $\boxed{\text{1.5} \times 10^{-5} \text{ M}}$.