Time taken by a $836\text{W}$ heater to heat one litre of water from $10^{\circ }\text{C}$ to $40^{\circ }\text{C}$ is:

The heat required to raise the temperature of water is given by:
$$Q=mc\Delta \theta$$
Given:Volume of water $=1\text{litre}⟹$ mass $m=1\text{kg}=1000\text{g}$.Specific heat of water $c=4.2{\text{J/g}}^{\circ }\text{C}$.$\Delta \theta =40^{\circ }\text{C}-10^{\circ }\text{C}=30^{\circ }\text{C}$.
$$Q=1000\times 4.2\times 30=126000\text{J}$$
The energy supplied by the heater in time $t$ is $E=P\times t$.
$$836\times t=126000$$
$$t=\frac{126000}{836}\approx 150.7\text{s}$$
Rounding to the nearest option, the time is $150\text{s}$.