A direct current of 4   A and an alternating current of peak value 4   A flow through resistance…

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  $H_2=i_{\mathrm{rms}}^2{R}_{2}t \left\{i_{\mathrm{rms}}=\frac{i_0}{\sqrt{2}}\right\}$  

$H_2=\frac{i_0^2}{2}{R}_{2}t$

$H_2=\frac{4^2}{2}\times \left(2\right)t=16t$

Thus, the ratio is $H_1:H_2=3:1$