How much of Sodium dichloro-s-triazinetrione dihydrate should I use to replace bleach?
According to a relevant reference (Ref.1) Following reactions happen when sodium hypochlorite and sodium dichloro-s-triazinetrione dissolve in water, respectively:

$$\ce{NaOCl + H2O -&gt; HOCl + NaOH} \tag {1}$$

$$\ce{NaCl2(NCO)3 + 2 H2O &lt;=&gt; 2 HOCl + NaH2(NCO)3} \tag {1}$$

Hence it is safe to assume that you need only $\pu{0.5 mol}$ of $\ce{NaCl2(NCO)3}$ (sodium dichloro-s-triazinetrione) or in your case $\ce{NaCl2(NCO)3.2H2O}$ (sodium dichloro-s-triazinetrione dihydrate), in place of each $\pu{1.0 mol}$ of $\ce{NaOCl}$ used. Let's see how much of $\ce{NaOCl}$ in $\pu{1.0 US gal}$ of $5.25\%$ bleach solution:

$$5.25\%(w/v) = \\ 
 \frac{\pu{5.25 g} \;\ce{NaOCl}}{\pu{100 mL} \;\text{solution}}\times \frac{\pu{1.0 mol} \;\ce{NaOCl}}{\pu{74.44 g} \;\ce{NaOCl}}\times \frac{\pu{1000 mL} \;\text{solution}}{\pu{1.0 L} \;\text{solution}}\times \frac{\pu{3.79 L} \;\text{solution}}{\pu{1.00 US gal} \;\text{solution}}\\
 = \pu{2.673 mol/US gal}$$

As states above, you need only half of this amounts in solid $\ce{NaCl2(NCO)3.2H2O}$. You can calculate the amount of $\ce{NaCl2(NCO)3.2H2O}$ in grams as follows:

$$\pu{1 US gal} \: \text{of } 5.25\%(w/v)  = \\
\pu{2.673 mol}  \; \text{of } \ce{NaOCl} \times \frac{\pu{0.5 mol} \; \ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \; \ce{NaOCl}} \times \frac{\pu{255.98 g} \;\ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \;\ce{NaCl2(NCO)3.2H2O}} \\
= \pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}$$

Thus, you can dissolve $\pu{342.1 g}$ of $\ce{NaCl2(NCO)3.2H2O}$ in your 30-gal (US) water barrel to get the same effect of 1-US gal of 5.25% bleach solution. Keep in mind that rate of chlorine release from $\ce{NaCl2(NCO)3.2H2O}$  is slower than that of bleach solution (Ref.1).



Late Edit:

This is about comment made by OP on my post. In that comment, OP states that the cost for dichloro-s-triazinetrione dihydrate (Di-Chlor) and beach are almost same. But I do not agree with that. Just for curiosity, I checked the prices for two chemicals: You can buy $\pu{50 lbs}$ bucket of Di-Chlor granules [online] from Pool Supply World for $\pu{132.99 USD}$; and You can buy $\pu{121 fl \: OZ}$ container of 6% bleach solution from Target for $\pu{4.49 USD}$ (let's assume its about $\pu{1 US\:Gal}$ of 5.25% bleach solution). According to my calculations above, $\pu{50 lbs}$ bucket of Di-Chlor granules equal to:

$$\pu{50.0 lbs} \; \ce{NaCl2(NCO)3.2H2O} \times \frac {\pu{454 g}}{\pu{1.0 lbs}}  \times \frac{\pu{1.0 US\:Gal}  \;\text{5.25% }\ce{NaClO}}{\pu{342.1 g}  \;\ce{NaCl2(NCO)3.2H2O}}\\
 \approx  \pu{66.5 US\:Gal} \text{ of 5.25% }\ce{NaClO} \text{ solution} $$

As a consequence, the cost for $\pu{50 lbs}$ bucket of Di-Chlor granules is $\pu{132.99 USD}$ while the cost for $\pu{66 US\:Gal}$ of 5.25% bleach solution  containers is $\pu{4.49 USD} \times 66  \approx \pu{296.34 USD}$ (more than twice the cost for Di-Chlor granules needed).



References:


Thomas Clasen, Paul Edmondson, "Sodium dichloroisocyanurate (NaDCC) tablets as an alternative to sodium hypochlorite for the routine treatment of drinking water at the household level," International Journal of Hygiene and Environmental Health 2006, 209(2), 173-181 (https://doi.org/10.1016/j.ijheh.2005.11.004).