A

Step 1: Identify the equation for pressure as a function of volume given in the problem, which is \( P = kV^3 \).
Step 2: To find the work done \( W \) during the thermodynamic process, we use the formula for work done on a gas, given by the integral of pressure with respect to volume:
\( W = \int_{V_1}^{V_2} P \, dV \).
Step 3: Substituting the expression for pressure into the formula:
\( W = \int_{V_1}^{V_2} kV^3 \, dV \).
Step 4: Calculate the integral:
\( W = k \left[ \frac{V^4}{4} \right]_{V_1}^{V_2} = k \left( \frac{V_2^4}{4} - \frac{V_1^4}{4} \right) = \frac{k}{4} (V_2^4 - V_1^4) \).
Step 5: The change in temperature suggests that the process also changes the internal energy of the gas, and we can use the ideal gas law: \( PV = nRT \).
Step 6: Since the question involves a change in temperature (from \( 100^\circ C \) to \( 300^\circ C \)), we calculate the work done across this range of temperature:
\( W = nR \Delta T = nR(300-100) = nR(200) \).
Step 7: Therefore, we can identify the work done as: \( W = \frac{k}{4} (V_2^4 - V_1^4) \) in terms of the moles of gas (n) and R.
Thus, the computed expression for work done will yield option A.