A direct current of 4A and an alternating current of peak value 4A...

Heat produced in resistor when direct current flows is ${H}_{1} = {i}^{2}{R}_{1}t$

${H}_{1} = {4}^{2}\left(3\right)t = 16\left(3\right)t$

Heat produced in resistor when alternating current flows is ${H}_{2} = {i}_{rms}^{2}{R}_{2}t\left\{{i}_{rms} = \dfrac{{i}_{0}}{\sqrt{2}}\right\}$

${H}_{2} = {i}_{0}^{2}2{R}_{2}t$

${H}_{2} = \dfrac{{4}^{2}}{2} \times \left(2\right)t = 16t$

Thus, the ratio is ${H}_{1}: {H}_{2} = 3: 1$