The work done by the system in a cyclic process involving one mole of an ideal monoatomic gas is – 50kJ per cycle. The heat absorbed by the system per cycle is

 As the process is cyclic, the change in internal energy will be zero.

So, according to the first law of thermodynamics:

$\mathrm{\Delta E}=\mathrm{q}+\mathrm{w}$

$$\begin{gathered} \mathrm{q}=-\mathrm{w} \\ \mathrm{q}=-(-50) \\ \mathrm{q}=+50\text{ KJ} \end{gathered}$$

Therefore, the correct option is B.