Is there a small enough planet or asteroid you can orbit by jumping? I just had this idea of orbiting a planet just by jumping and then flying upon it on its orbit kind of like superman. So,
Would it be theoretically possible or is there a chance of that small body to be & remain its unity?

 A: Let's assume mass of the person plus spacesuit to be $m_1$=100kg
Asteroid density: $\rho=$2g/cm$^3$ (source) that is 2 000kg/m$^3$
15km/hour is a good common run. That's roughly v=4m/s
The orbital height is negligible comparing to the radius, assume 0 over surface.
Linear to angular velocity (1):
$$
\omega = {v \over r }
$$
Centripetal force (2):
$$
F = m r \omega ^2  
$$
Gravity force (3):
$$
F= G \frac{m_1 m_2}{r^2}
$$
Volume of a sphere (4):
$$
V = \frac{4}{3}\pi r^3 
$$
Mass of a sphere (5):
$$
m_2 = V \rho = \frac{4}{3}\pi r^3 \rho
$$
Combining (1),(2),(3), reducing:
$$
{ m_1 r v^2 \over r^2 } = G { m_1 * m_2 \over r^2 }
$$
$$
 r v^2 = G m_2 
$$
Combining with (5)
$$
 r v^2 = G \frac{4}{3}\pi r^3 \rho
$$
$$
 r^2 = \frac{v^2}{\rho G \frac{4}{3}\pi} 
$$
$$
 r = v ({\frac{4}{3}\pi G \rho})^{-{1 \over 2}}
$$
Substituting values:
$$
 r = 4 ({1.33333*3.14159* 6.67300*10^{-11} * 2000})^{-{1 \over 2}}  
$$
That computes to roughly 5.3 kilometers
More interestingly, the radius is directly proportional to the velocity,
$$
r[m] = 1337[s] * v [m/s] = 371.51[h/1000] * v[km/h] = 597[m*h/mile] * v[mph]
$$
So, a good walk on a 2km radius asteroid will get you orbiting.
Something to fit your bill would be Cruithne, a viable target for a space mission thanks to a very friendly orbit.
Note, while in rest on Cruithne, the astronaut matching the m_1=100kg would be pulled down with force of 4.5N while not in motion. That is like weighing about 450g or 1lbs on Earth.
A: If you want an idea what this might actually be like have a look at Kerbal Space Program. This is a game currently in development by Squad. So not real life, but the orbital physics is accurately modelled (atmospheric flight not so much, yet). There are several small moons and asteroids in the Kerbin system where you can do essentially this jump to orbit maneuvre using nothing but EVA suit thrusters. You can see examples in some of Scott Manley's videos. Here is a video featuring an interplanetary trip with an EVA suit - a 49 day space walk!
(I'm not affiliated with KSP, Squad or Scott Manley in any way, and since the question has been properly answered already I thought this might just be a fun thing to share. Also, KSP and the similar game Orbiter are good ways of building intuition for orbital mechanics. :) Hope this doesn't break the rules.)
A: No, not by jumping. Jumping gives you an acceleration only from the location on the surface. As soon as you leave the surface, you have no way of adjusting your orbit. Either you reach escape velocity, or you will return to your initial location after exactly one orbit.

The only way to prevent this would be to have an additional acceleration once you have departed from the surface. Spacecraft use rockets to do this. A tiny acceleration may be enough — though I wouldn't like approaching a planet with high speed only to move 5 cm over its surface with high speed!
Edit: A different way would be jump from a ladder, as Claudius pointed out in the other answer.
A: OK, I tried to do the math here. Something remotely resembling maths, at least.
Assumptions:


*

*It is possible to reach an orbital/horizontal speed of $v_O = 5\textrm{ ms}^{-1}$, for example by running.

*The density of the object to orbit is similar to Earth's density, i.e. $\rho = 5500\textrm{ kgm}^{-3}$.

*We want to orbit at a height of $2\textrm{ m}$ above the ground. You can get there with a ladder (Yes, you will have to start running on that ladder or something like that....how about stilts?).

*No atmosphere or other source of friction.


Layout:
The basic idea is to link the orbital velocity $v_O$ to the radius $r$ of the object. The mass is given by $ M = \frac{4}{3} \pi r^3 \rho$ (God I hope I remembered this formula correctly).
Calculation:
We have
\begin{eqnarray}
& v_O & = \sqrt{\frac{G M}{r+2\textrm{ m}}} = 5\textrm{ ms}^{-1} \\
\Rightarrow & M & = \frac{25\frac{\textrm{m}^2}{\textrm{s}^2} \left( r + 2\textrm{ m} \right)}{G} \\
\Rightarrow & 25 \frac{\textrm{m}^2}{\textrm{s}^2} r + 50 \frac{\textrm{m}^3}{\textrm{s}^2} & = \frac{4}{3} \pi G r^3 5500 \frac{\textrm{kg}}{\textrm{m}^3}
\end{eqnarray}
which then should give us $r$. I used Mathematica for this because it is half past eleven in the evening and I don’t want to guess solutions to get a starting point for polynomial division, getting:
In:  Solve[-4/3 * Pi * 6.67384*10^(-11) * x^3 * 5500 + 25 x + 50 == 0, x]
Out: {{x -> -4031.33327417391}, {x -> -2.00000049201392}, {x -> 4033.33327466592}}

That is, if you found an asteroid of $r \approx 4\textrm{ km}$, your dream might come true. However, if it is mostly ice (rather than molten iron, which I imagine would be a pretty good reason to stay in orbit), you will have to correct the 5500 up there to the density of ice, say, 930, and would then need an asteroid of $r \approx 9.8\textrm{ km}$.
Note that the assumption that $m_{\textrm{Human}} \ll m_{\textrm{Object}}$, encoded in the expression for orbital velocity, is fulfilled relatively well in these cases (five orders of magnitude).
Nevertheless, feel free to point out mistakes :)
A: Since the calculations are already in others' answers, I'll just refer to this great, classic xkcd. Deimos and Phobos, the two small moons of Mars, match (or almost match) the criteria SF and Claudius derive.
As Munroe points out,

(The diagram is a representation of the gravity wells of both moons, represented by their height at constant Earth surface gravity.)
Based on that I think you really should be able to run yourself into orbit using a smallish ramp and a fire extinguisher to stabilize your orbit on the other side (to avoid the pitfall gerrit mentions).
Deimos is between 10 and 15 km across and its escape velocity is about 20 km/h. At low altitudes, and since circular-orbit velocities are lower by $\sqrt{2}$ than escape velocities, you'd need to run up to some 15 km/h to orbit. Thus you'd do about one lap every three hours, whizzing along this ~city sized object at about bicycle speeds.

On the other hand, it's unlikely that you will last very long in that orbit. The reason for this is that orbits are elliptical only around perfectly spherical planets, and any irregularities in the body you're orbiting will tend to perturb and even destabilize your orbit. Even on the Moon, low orbits are unstable and end up crashing into the surface, as was the fate of a subsatellite deployed during Apollo 16, which lasted only a month in orbit. With something as lumpy as the Martian moons, you would probably want to stay well away!
