Answer:
Approximately [tex]3.43\; {\rm m\cdot s^{-2}}[/tex], assuming that air resistance is negligible in both occasions, and that [tex]g_{\text{earth}} = 9.81\; {\rm m\cdot s^{-2}}[/tex] near the surface of the Earth.
Explanation:
Let [tex]x[/tex] denote the displacement of the marker. Let [tex]a[/tex] denote the acceleration of the marker. Let [tex]t[/tex] denote the time it takes for the marker to reach the ground.
Under the assumptions, acceleration of the marker would be constant, and the SUVAT equations would apply. Rearrange the SUVAT equation [tex]x = (1/2)\, a\, t^{2}[/tex] to find acceleration [tex]a[/tex]:
[tex]\begin{aligned}a &= \frac{2\, x}{t^{2}}\end{aligned}[/tex].
Let [tex]a_{\text{Mars}}[/tex] and [tex]t_{\text{Mars}}[/tex] denote the acceleration and time taken on Mars. Similarly, let [tex]a_{\text{Earth}}[/tex] and [tex]t_{\text{Earth}}[/tex] denote the acceleration and time taken on Earth. It is implied that the Marker travelled the same distance (same displacement, [tex]x[/tex]) both on Earth and on Mars.
Using the SUVAT equation from above:
[tex]\begin{aligned}a_{\text{Mars}} &= \frac{2\, x}{{t_{\text{Mars}}}^{2}}\end{aligned}[/tex].
[tex]\begin{aligned}a_{\text{Earth}} &= \frac{2\, x}{{t_{\text{Earth}}}^{2}}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{a_{\text{Mars}}}{a_{\text{Earth}}} &= \frac{\displaystyle \frac{2\, x}{{{t_{\text{Mars}}}^{2}}}}{\displaystyle \frac{2\, x}{{{t_{\text{Earth}}}^{2}}}} \end{aligned}[/tex].
[tex]\begin{aligned}\frac{a_{\text{Mars}}}{a_{\text{Earth}}} &= \left(\frac{{t_{\text{Earth}}}}{{t_{\text{Mars}}}}\right)^{2}\end{aligned}[/tex].
[tex]\begin{aligned}a_{\text{Mars}} &= \left(\frac{{t_{\text{Earth}}}}{{t_{\text{Mars}}}}\right)^{2}\, a_{\text{Earth}} \\ &= \left(\frac{1.3\; {\rm s}}{2.2\; {\rm s}}\right)^{2}\, (9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 3.43\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
