Applying boundary conditions, $120=100(1+200 \alpha)$

$\alpha=10^{-3}\, K^{-1}$

It is given that temperature increases at a constant rate from $300\, \mathrm{K}$ to $500\, \mathrm{K}$ in $30 \,\mathrm{s}$. Hence, $T(t)=300+20 t / 3$

By Joule's Law, heat dissipated in a resistor is given by:

$W=\int_{0}^{30} \frac{V^{2}}{R} d t$

$=\int_{0}^{30} \frac{V^{2}}{R_{0}\left(1+\alpha\left(T-T_{0}\right)\right)} d t$

$=\frac{V^{2}}{R_{o}} \int_{0}^{30} \frac{1}{(1+20 \alpha t / 3)} d t$

Solving, $W=400 \ln (6 / 5)$

Work done on resistor $=-W=400 \ln (5 / 6) \,J$