Given, mass of the particle, $m = 30 \text{ gm} = 30 \times 10^{-3} \text{ kg} = 0.03 \text{ kg}$. The position of the particle is given by $x = 3t - 4t^2 + t^3$. The velocity of the particle is the rate of change of position: $v = \frac{dx}{dt} = \frac{d}{dt}(3t - 4t^2 + t^3) = 3 - 8t + 3t^2$

Initial velocity at $t = 0 \text{ s}$: $v_i = 3 - 8(0) + 3(0)^2 = 3 \text{ m/s}$

Final velocity at $t = 4 \text{ s}$: $v_f = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19 \text{ m/s}$

According to the work-energy theorem, the work done by the net force is equal to the change in kinetic energy : $W = \Delta K = K_f - K_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$ $W = \frac{1}{2} m (v_f^2 - v_i^2)$ $W = \frac{1}{2} \times 0.03 \times (19^2 - 3^2)$ $W = 0.015 \times (361 - 9) = 0.015 \times 352 = 5.28 \text{ J}$

Thus, the work done during the first $4 \text{ seconds}$ is $5.28 \text{ J}$.