Answer
$19$
Work Step by Step
We need to find How many bright-dark-bright fringes are
shifted when the cell is filled with air.
We know that the gas increases the number of wavelengths in one arm of the interferometer where each extra wavelength will create one bright-dark-bright fringe shift.
Recalling that the number of the shifted fringes is given by
$$\Delta m=m_2-m_1$$
where $m_1=2d/\lambda_{\rm vaccum}$, and $m_2=2d/\lambda$ where $\lambda=\lambda_{\rm vaccum}/n_{\rm air}$.
So,
$$\Delta m=\dfrac{2dn_{\rm air}}{\lambda_{\rm vaccum}}-\dfrac{2d}{\lambda_{\rm vaccum}}=\dfrac{2d(n_{\rm air}-1)}{\lambda_{\rm vaccum}}$$
Plugging the known;
$$\Delta m= \dfrac{2(0.02)(1.00028-1)}{600\times 10^{-9}}=\bf 18.67$$
$$\Delta m= \color{red}{\bf 19}$$
Hence, we will get about 19 shifted bright-dark-bright fringe.
