This is how you do the calculation.
The elapsed time on an observer's clock is called the proper time, $\tau$, and it is calculated by integrating the metric:
$$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - \frac{dr^2}{1-\frac{2GM}{c^2r}} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 $$
In this case we'll assume all motion is radial so $d\theta = d\phi = 0$ and the equation simplifies to:
$$ d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)dt^2 - \frac{1}{c^2}\frac{dr^2}{1-\frac{2GM}{c^2r}} \tag{1} $$
The coordinates $t$ and $r$ are the Schwarzschild time and radial coordinates i.e. the time and radial coordinates for an observer far from the Earth. For the third clock that stays on Earth the calculation is easy since $dr = 0$ (i.e. it isn't moving) and equation (1) gives:
$$ d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)dt^2 $$
And this immediately integrates to:
$$ \tau_3 = \sqrt{1-\frac{2GM}{c^2r}}\,t \tag{2} $$
which is just the equation for time dilation in the gravitational field outside a spherically symmetric mass.
For the first clock we'll make two approximations:
the time spent near the Earth is small compared to the total travel time
we'll ignore the turnaround time at the far end of the trip
So in effect the third observer is moving at constant velocity $v$ at large $r$ where $2GM/c^2r$ is effectively zero. For the outbound journey the distance is simply given by $r = vt$ so $dr = vdt$, and substituting in equation (1) gives:
$$ d\tau^2 = dt^2 - \frac{v^2}{c^2}dr^2 $$
and this immediately integrates to:
$$ \tau_\text{out} = \sqrt{1 - \frac{v^2}{c^2}}\,\frac{t}{2} $$
where the time is $t/2$ because we're only considering half the journey. Since the journey is symmetric the outboud and return times are the same, so the total time for clock 1 is:
$$ \tau_1 = 2\tau_\text{out} = \sqrt{1 - \frac{v^2}{c^2}}\,t \tag{3} $$
which is just the expression for time dilation in special relativity.
The calculation for the second clock is really hard. We'll once again assume we can ignore the Earth's gravitational field since the rocket spends only a short time near the Earth. In that case the distance as a function of time is given by:
$$ r = \frac{c^2}{a} \left(\sqrt{1 + \frac{at}{c^2}} - 1\right) $$
where $a$ is the constant acceleration as measured on the rocket. Using this to substitute for $dr$ in equation (1) is somewhat involved so I'll cheat and just quote the result from Phil Gibbs' article on the relativistic rocket. For the first quarter of the journey we get:
$$ \tau_\text{qtr} = \frac{c}{a} \text{arcsinh}\left(\frac{a}{c}\frac{t}{4}\right) $$
Again assuming symmetry each quarter of the journey takes the same time, so the total time for the second clock is:
$$ \tau_2 = 4\frac{c}{a} \text{arcsinh}\left(\frac{a}{c}\frac{t}{4}\right) \tag{4} $$
I'll leave the last stage of the calculation to you. Equations (2), (3) and (4) give you the three times you need for the comparison.
