Answer
$8.04\times 10^{-8}\;\rm s$
Work Step by Step
The time the plane would take as seen by a stationary observer on Earth is given by
$$v=\dfrac{2\pi R_E}{\Delta t_E}$$
where $v$ is the plane's speed, $R_E$ is the Earth's radius on the equator, and $\Delta t_E$ is the measured time by Earth's observer.
Thus,
$$\Delta t_E=\dfrac{2\pi R_E}{v}\tag 1$$
And the time measured on the plane is given by
$$\Delta t_{plane}=\dfrac{\Delta t_E}{\gamma}=\Delta t_E\sqrt{1-\dfrac{v^2}{c^2}}$$
Using the binomial expansion as the author told us.
Since $\dfrac{v^2}{c^2}\lt \lt 1$, so
$$\sqrt{1-\frac{v_{\rm equator }^2}{c^2}}=1+\frac{1}{2}\dfrac{v^2}{c^2}$$
Thus,
$$\Delta t_{plane} =\Delta t_E\left[1+ \dfrac{v^2}{2c^2}\right]$$
Plugging from (1);
$$\Delta t_{plane} =\dfrac{2\pi R_E}{v}\left[1+ \dfrac{v^2}{2c^2}\right]$$
$$\Delta t_{plane} =\dfrac{2\pi R_E}{v}\left[1+ \dfrac{v^2}{2c^2}\right]\tag 2$$
So that the time difference between the two frames is given by
$$\Delta t=(\Delta t_E-\Delta t_{plane})$$
Plugging from (1) and (2);
$$\Delta t=\dfrac{2\pi R_E}{v}- \left(\dfrac{2\pi R_E}{v}\left[1+ \dfrac{v^2}{2c^2}\right]\right)$$
$$\Delta t=\bigg|\dfrac{2\pi R_E}{v}\left(1- \left[1+ \dfrac{v^2}{2c^2}\right]\right)\bigg|$$
$$\Delta t=\dfrac{2\pi R_E}{v}\left( \dfrac{v^2}{2c^2} \right)$$
$$\Delta t= 2\pi R_E \left( \dfrac{v }{2c^2} \right)$$
Plugging the known, and remember to convert the speed of the plane to SI units.
$$\Delta t= 2\pi \times 6.38\times10^6\left( \dfrac{\frac{ 1300\times 1000}{60^2} }{2(3\times10^8)^2} \right)$$
$$\Delta t= \color{red}{\bf 8.04\times 10^{-8}}\;\rm s$$
