To solve this problem, we need to find the total heat produced in the resistance $R. \\$The charge $Q$ flowing through the resistance is given by $Q = at - bt^2. \\$The current $I$ is the rate of change of charge with respect to time:$\\ I = \frac{dQ}{dt} = \frac{d}{dt}(at - bt^2) \\$Differentiate $Q$ with respect to $t: \\ I = a - 2bt \\$The power $P$ dissipated in the resistance $R$ is given by $P = I^2 R. \\$Substitute the expression for $I: \\ P = (a - 2bt)^2 R \\$To find the total heat $H$ produced, integrate the power over time:$\\ H = \int P \, dt = \int (a - 2bt)^2 R \, dt \\$Expand the integrand:$\\ (a - 2bt)^2 = a^2 - 4abt + 4b^2t^2 \\$Substitute back into the integral:$\\ H = R \int (a^2 - 4abt + 4b^2t^2) \, dt \\$Integrate term by term:$\\ H = R \left[ \int a^2 \, dt - \int 4abt \, dt + \int 4b^2t^2 \, dt \right] \\$Calculate each integral:$\\ \int a^2 \, dt = a^2 t \\ \int 4abt \, dt = 4ab \frac{t^2}{2} = 2abt^2 \\ \int 4b^2t^2 \, dt = 4b^2 \frac{t^3}{3} = \frac{4b^2 t^3}{3} \\$Combine the results:$\\ H = R \left[ a^2 t - 2abt^2 + \frac{4b^2 t^3}{3} \right] \\$To find the total heat, evaluate the integral from $t = 0$ to $t = \frac{a}{2b}$ (when $I = 0): \\ H = R \left[ a^2 \left(\frac{a}{2b}\right) - 2ab \left(\frac{a}{2b}\right)^2 + \frac{4b^2}{3} \left(\frac{a}{2b}\right)^3 \right] \\$Simplify each term:$\\ a^2 \left(\frac{a}{2b}\right) = \frac{a^3}{2b} \\ 2ab \left(\frac{a}{2b}\right)^2 = \frac{2ab \cdot a^2}{4b^2} = \frac{a^3}{2b} \\ \frac{4b^2}{3} \left(\frac{a}{2b}\right)^3 = \frac{4b^2 \cdot a^3}{24b^3} = \frac{a^3}{6b} \\$Combine the terms:$\\ H = R \left[ \frac{a^3}{2b} - \frac{a^3}{2b} + \frac{a^3}{6b} \right] \\ H = R \left[ 0 + \frac{a^3}{6b} \right] \\ H = \frac{a^3 R}{6b} \\$Therefore, the total heat produced in the resistance $R$ is $\frac{a^3 R}{6b}. \\$This corresponds to Option 3.