Answer
$0.569\;\rm m$
Work Step by Step
First of all, we need to find the angle of the first order of the two wavelengths.
We know, for bright fringes in a diffraction grating, that
$$m\lambda=d\sin\theta $$
$$\theta=\sin ^{-1}\left(\dfrac{m\lambda}{d}\right)\tag 1 $$
And we know that the distance from the center fringe to the first-order bright fringe is given by
$$x=L\tan\theta $$
Plugging from (1);
$$x=L\tan\left[\sin ^{-1}\left(\dfrac{m\lambda}{d}\right)\right] $$
So, for the first order for the first wavelength;
$$x_1=L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_1}{d}\right)\right] $$
and for the first order for the second wavelength;
$$x_2=L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_2}{d}\right)\right] $$
Therefore, the linear distance between the first-order bright fringes of these two wavelengths on the screen is given by
$$\Delta x=x_2-x_1$$
$$\Delta x=L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_2}{d}\right)\right]-L\tan\left[\sin ^{-1}\left(\dfrac{ \lambda_1}{d}\right)\right] $$
Plugging the known;
$$\Delta x=2.5\tan\left[\sin ^{-1}\left(\dfrac{ 6.8\times10^{-7}}{\dfrac{1}{7200}\times10^{-2}}\right)\right]-2.5\tan\left[\sin ^{-1}\left(\dfrac{ 4.4\times10^{-7}}{\dfrac{1}{7200}\times10^{-2}}\right)\right] $$
$$\Delta x=\color{red}{\bf 0.569}\;\rm cm$$