The useful heat absorbed by the water is:

$$Q = m c \Delta T$$

The energy delivered by the stove is $P t$, of which only the fraction $\eta$ is useful:

$$\eta P t = m c \Delta T$$

Solving for $t$:

$$t = \frac{m c \Delta T}{\eta P}$$

Substituting $m = 2$ kg, $c = 4.2 \times 10^{3}$ J/(kg$\cdot{}^\circ$C), $\Delta T = 80\,^\circ\text{C}$, $\eta = 0.30$, $P = 2000$ W:

$$t = \frac{2 \times 4.2 \times 10^{3} \times 80}{0.30 \times 2000} = \frac{6.72 \times 10^{5}}{600} = 1120 \text{ s} \approx 18.7 \text{ min}$$