Should I use specific heat to calculate how much fuel is required?
$$\Delta H = mc_p\Delta T$$

Where $\Delta H$ is either energy absorbed or released at constant pressure.
Assuming this specific problem refers to $3\ \mathrm{kg}$ of water:
$$
\Delta H = \left(3000\ \mathrm g\right)\left(4.184\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\right)\left(80\ \mathrm K\right)
$$

That means that the energy required is: $1\,004\,160$ joules.

Now, simply divide those 1 004 160 joules by the 36 000 joules your fuel releases per gram.
$$
m_\text{fuel}= \frac{1\,004\,160\ \mathrm J}{36\,000\ \mathrm{J\cdot g^{-1}}} = 27.89\ \mathrm g
$$