Answer
The time it takes is $33.3s$
Work Step by Step
1) When engine 1 and engine 2 are fired in the same direction: $$\sum \vec{F}=\vec{F_1}+\vec{F_2}$$ $$\sum F=F_1+F_2=2F$$
We have the probe's initial velocity $v_0=0$, its acceleration $a$, the distance it travels $\Delta x$ and the time to travel it $t_1=28s$.
$$\Delta x=v_0t+\frac{1}{2}at_1^2$$ $$\Delta x=\frac{1}{2}at^2_1$$
From Newton's 2nd Law, we can rewrite $a$: $$\Delta x=\frac{1}{2}\frac{\sum F}{m_p}t^2_1=\frac{2Ft^2_1}{2m_p}=\frac{Ft^2_1}{m_p}$$
2) When engine 1 and engine 2 are fired in perpendicular direction: $$\sum \vec{F}=\vec{F_1}+\vec{F_2}$$ $$\sum F=\sqrt{F_1^2+F_2^2}=\sqrt{F^2+F^2}=F\sqrt2$$
Similarly, we follow the calculations from 1) for $\Delta x$ and get
$$\Delta x=\frac{F\sqrt2t_2^2}{2m_p}$$
Therefore, $$\frac{F\sqrt2t_2^2}{2m_p}=\frac{Ft^2_1}{m_p}$$ $$\frac{t^2_2\sqrt2}{2}=t^2_1$$ $$t_2^2=\sqrt2 t_1^2$$ $$t_2=\sqrt[4]2t_1=\sqrt[4]2\times28=33.3s$$
