Answer: 40%

For an isobaric process, $Q = nC_P \Delta T$ and $W = nR \Delta T$

Now, $C_P = \large\frac{\gamma R}{ \gamma -1}$ for an isobaric process, where $\gamma$ is the adiabatic index or the heat capacity ratio.

$\Rightarrow \large\frac{\Delta W}{\Delta Q}$$ = \large\frac{R}{C_P}$$ =\large\frac{\gamma -1}{\gamma}$

For monatomic gases, $\gamma = \large\frac{5}{3}$

$\Rightarrow \large\frac{\Delta W}{\Delta Q}$$ =\large\frac{\gamma -1}{\gamma}$$ = \Large\frac{\frac{5}{3}-1}{\frac{5}{3}}$$ = 0.4 = 40\%$

This is the fraction of heat supplied for external work.