To solve this problem, we need to calculate the work done by the force $F$ as the particle moves from $y = 0$ to $y = 1$ m.$\\$The force acting on the particle is given by:$\\ F = 20 + 10y \\$The work done by a variable force is given by the integral:$\\ W = \int_{y_1}^{y_2} F \, dy \\$Substitute the given force into the integral:$\\ W = \int_{0}^{1} (20 + 10y) \, dy \\$Separate the integral into two parts:$\\ W = \int_{0}^{1} 20 \, dy + \int_{0}^{1} 10y \, dy \\$Calculate each integral separately:$\\ \int_{0}^{1} 20 \, dy = 20y \bigg|_{0}^{1} = 20(1) - 20(0) = 20 \\ \int_{0}^{1} 10y \, dy = 10 \left(\frac{y^2}{2}\right) \bigg|_{0}^{1} = 10 \left(\frac{1^2}{2}\right) - 10 \left(\frac{0^2}{2}\right) = 5 \\$Add the results of the two integrals to find the total work done:$\\ W = 20 + 5 = 25$ J$\\$Therefore, the work done by the force is $25$ J. This corresponds to Option 3.