Answer
$0.24$
Work Step by Step
We know that the thermal efficiency for the Brayton cycle is given by
$$\eta=1-\left[\dfrac{P_2}{P_1}\right]^\frac{1-\gamma}{\gamma}\tag 1$$
We also know, for the Brayton cycle, there 4 processes two adiabatic processes and two thermal processes.
Henev, for adiabatic process,
$$P_1V_1^\gamma=P_2V_2^\gamma$$
Thus,
$$\dfrac{P_2}{P_1}=\left[\dfrac{V_1}{V_2}\right]^\gamma$$
Plugging into (1);
$$\eta=1-\left[\left[\dfrac{V_1}{V_2}\right]^\gamma\right]^\frac{1-\gamma}{\gamma}$$
Recalling that $V_2=\frac{1}{2}V_1$, so that
$$\eta=1-\left[ \dfrac{V_1}{\frac{1}{2}V_1} \right]^{1-\gamma} $$
$$\eta=1-\left[2 \right]^{1-\gamma}$$
where $\gamma=1.4$ for diatomic gases.
$$\eta=1-\left[2 \right]^{1-1.4}$$
$$\eta=\color{red}{\bf 0.24}$$
