The current is given as,

$I=\frac{d Q}{d t}$

$=2-16 t$

The heat produced in the resistor is given as,

$H=\int I^{2} R d t$

$=\int_{0}^{\frac{1}{8}}(2-16 t)^{2} R d t$

$=\left[4 t+\frac{256 t^{3}}{3}-32 t^{2}\right]_{0}^{\frac{1}{8}} R$

$=\frac{R}{6} \mathrm{J}$

Thus, the total heat produced is $\frac{R}{6} \mathrm{J}$