To solve this problem, we need to determine the power developed by the force at time $t. \\$Power $P$ is given by the dot product of force $F$ and velocity $v: \\ P = F \cdot v \\$Given the force $F = (2t \hat{i} + 3t^2 \hat{j})$ N, we first need to find the velocity $v. \\$The acceleration $a$ is given by $a = \frac{F}{m},$ where $m = 1$ kg.$\\ a = (2t \hat{i} + 3t^2 \hat{j})$ m/s$^2 \\$Since the body starts from rest, the initial velocity $v_0 = 0. \\$The velocity $v$ at time $t$ is the integral of acceleration:$\\ v = \int a \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt \\$Integrating each component separately:$\\ v = \left( \int 2t \, dt \right) \hat{i} + \left( \int 3t^2 \, dt \right) \hat{j} \\$Calculate the integrals:$\\ \int 2t \, dt = t^2 + C_1 \\ \int 3t^2 \, dt = t^3 + C_2 \\$Since the initial velocity is zero, $C_1 = 0$ and $C_2 = 0. \\ v = (t^2 \hat{i} + t^3 \hat{j})$ m/s$\\$Now, calculate the power $P: \\ P = F \cdot v = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) \\$Perform the dot product:$\\ P = (2t \cdot t^2) + (3t^2 \cdot t^3) \\ P = 2t^3 + 3t^5 \\$Therefore, the power developed by the force at time $t$ is $(2t^3 + 3t^5)$ W.$\\$This corresponds to Option 2.