Wiki and lots of sources all say the same thing. "1 part of 3% hydrogen peroxide to 10 parts water."
When talking about aqueous solutions, the concentrations can be expressed as percent by weights, $\% (w/w)$, or by volume, $\% (v/v)$, or by weight in parts of solute by weight in 100 part of solution by volume, $\% (w/v)$. The above statement in the question means if you take 100 part of the solution by weight (e.g., $\pu{100 g}$ of solution), it contains 3 part of hydrogen peroxide and 97 parts of water by weights $\left(3\% (w/w)\right)$. It also can mean that it may contain 3 part of hydrogen peroxide to 97 parts of water by volumes $\left(3\% (v/v)\right)$ if the lebel doesn't say it is whether $\% (w/w)$ or $\% (v/v)$. The third kind is 3 part of hydrogen peroxide by weight to 100 parts of solution by volume $\left(3\% (w/v)\right)$, which is common for prepared chemical reagents to use in the laboratories.
Commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$. However, when prepared as a chemical reagent, it is $\% (w/v)$. That means $\pu{100 mL}$ of $3\% (w/v)$ $\ce{H2O2}$ solution has $\pu{3 g}$ or the molarity of the solution:
$$3\% (w/v) = \frac{\pu{3 g} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.882 mol L-1}$$
On the other hand, since commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$, and density of pure $\ce{H2O2}$ is $\pu{1.450 g mL-1}$ at $\pu{20 ^\circ C}$ (Wikipedia):
$$3\% (v/v) = \frac{\pu{3 mL} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 mL}\cdot \pu{1.45 g mL-1} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{43.5 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{1.28 mol L-1}$$
Household bleach is aqueous $\ce{NaClO}$ $(\pu{74.44 g mol-1})$ solution, sometimes included certain percent of $\ce{NaOH}$. If you have $3\%$ solution, you can do the same calculations as above to find out the correct molarity if the solution is given whether $3\% (w/v)$ or $3\% (v/v)$ (and it is without $\ce{NaOH}$). Usually, it is always given in $\% (w/v)$. Sometimes, the default densities may be provided instead of percentages by the manufacturer, which may differ for the different manufacturer since the manufacturer may change the density by adding salt or other chemicals to the solution. Therefore, it is hard to calculate molarity using provided default density if you don't know what else other than $\ce{NaClO}$.
Let's consider the solution you have is $3\% (w/v)$ $\ce{NaClO}$:
$$3\% (w/v) = \frac{\pu{3 g} \ \ce{NaOCl}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{NaOCl}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{NaOCl}}{\pu{74.44 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.403 mol L-1}$$
The reaction is:
$$\ce{NaClO + H2O2 -> NaCl + H2O + O2}$$
Accordingly, $\pu{1.0 mol}$ of $\ce{NaOCl}$ will react with $\pu{1.0 mol}$ of $\ce{H2O2}$. Therefore, the volume $(V_\ce{H2O2})$ of $3\% (w/v)$ $\ce{H2O2}$ needed is react completely with $\pu{1.0 L}$ of $3\% (w/v)$ $\ce{NaOCl}$ can be calculated using $M_1V_1 = M_2V_2$ as follows:
$$M_\ce{H2O2}V_\ce{H2O2} = M_\ce{NaOCl}V_\ce{NaOCl} \ \Rightarrow \ V_\ce{H2O2} = \frac{M_\ce{NaOCl}V_\ce{NaOCl}}{M_\ce{H2O2}} \\ = \frac{\pu{0.403 mol L-1}\cdot \pu{1.0 L}}{\pu{0.882 mol L-1}} = \pu{0.457 L}$$
Therefore, since you have one US gallon of bleach, which equals to $\pu{3.785 L}$, you need $0.457 \times 3.785 = \pu{1.73 L}$ of $3\% (w/v)$ $\ce{H2O2}$ solution.
Your final solution is $\frac{0.403}{1.457} = \pu{0.277 mol L-1}$ $\ce{NaCl}$ solution (of course, only if manufacturers did not add additives).
