What ratio of 73% calcium hypochlorite powder to water mimics the strength of sodium hypochlorite (bleach)?
To start, we must first understand that we can ignore the counterions to the $\ce{Ca(ClO)2}$ and the $\ce{NaClO}$. Typical household bleach claims to have a concentration of 6% (w/v). Now we can do some basic arithmetic to convert the percent weight by volume of sodium hypochlorite in water to a molar concentration:

$\mathrm{\frac{0.06\,g}{1\,mL}}$$\mathrm{\frac{1\,mol}{74.44\,g}}$$\mathrm{\frac{1000\,mL}{1\,L}}$$\mathrm{\,= 0.8060\,M \,OCl^-}$

The first fraction in the above equation is based on the definition of w/v, which is the number of grams of a substance dissolved in one milliliter of a solution times 100%. In the next fraction, we convert the grams $\ce{NaClO}$ into moles of $\ce{NaClO}$. Finally, in the last fraction, we convert milliliters into liters to give the number of moles per liter of $\ce{NaClO}$, or simply $\ce{ClO^-}$ without the counter ion. 

Now that we know the molar concentration of $\ce{ClO^-}$ we are aiming for, we can convert this into the number of grams of pure $\ce{Ca(ClO)2}$ we would need to to achieve this concentration in one gallon of water

$\mathrm{\frac{0.8060\,mol}{1\,L}}$$\mathrm{\frac{1\,mol}{2\,mol}}$$\mathrm{\frac{142.98\,g}{1\,mol}}$$\mathrm{\frac{3.78541\,L}{1\,gal}}$$\mathrm{\,=\frac{218.1\,g\,OCl^-}{gal}}$

In the second fraction, we converted the moles of $\ce{ClO^-}$ into the number of moles of $\ce{Ca(ClO)2}$ (as there is only one unit of $\ce{Ca(ClO)2}$ per two units of $\ce{ClO^-}$). In the third fraction, we converted the number of moles of $\ce{Ca(ClO)2}$ into grams. In the last fraction, we converted the number of liters into the number of gallons. 

As you stated, the powder only contains 73% $\ce{Ca(ClO)2}$ by mass. We know we need a total of 218.1 g of $\ce{Ca(ClO)2}$, so we can use the following conversion factor:

$\mathrm{\frac{218.1\,g\,OCl^-}{1\,gal}}$$\mathrm{\frac{1\,g}{0.73\,g}}$$\mathrm{\,=298.8\,g\,powder}$ (Rounded to 0.3 kg since we only have one significant figure)

In the second fraction, we know that per every gram of your powder, 73 percent (0.73 g) of the $\ce{Ca(ClO)2}$ will be present. Thus, the conversion factor is valid. 

Final Answer: 0.3 kg of Powder