Why don't stationary charge feel force from a current carrying wire? The current carrying wire doesn't apply any magnetic force on nearby charge $q$( positive stationary charge) because it has 0 velocity in lab frame. We found that there is no force on q by wire. But if we take a moving frame then, q is in relative motion and hence a current carrying wire applies a magnetic force on q. Let's denote this magnetic force in moving frame by $F_B$. Since net force on charge is still 0 there is some force needed  to cancel out this $F_B$ force.
This was answered by length contraction. I have seen many videos referring to this as solution but i don't think if it works. I need a calculation which can show how length contraction actually helps in cancelling out $F_B$.
For calculation you are going to do i would like to refer some sign but you can take numerical value if you wish.
Area of cross-section of wire, $A$; length of wire in lab frame, $L$; electron density of wire in lab frame, $n$ ; the average velocity of electrons in lab frame, $v_d$; the moving frame is moving in opposite direction of electrons motion as seen from lab frame and the velocity of this frame relative to lab frame is $v_F$. The charge q is placed at $r$ distance from center of wire that. Moving frame is parallel to straight wire.
 A: 
I need a calculation which can show how length contraction actually helps in cancelling out FB.

The answer by @Frobenius shows the calculation using ordinary vectors. Since the question asks about length contraction, it is a question about relativity. So this answer shows the calculations using the relativistic framework of four-vectors and tensors.
The electric and magnetic fields are combined into the antisymmetric electromagnetic field tensor $$F^{\mu \nu }=\left(
\begin{array}{cccc}
 0 & -E_x & -E_y & -E_z \\
 E_x & 0 & -B_z & B_y \\
 E_y & B_z & 0 & -B_x \\
 E_z & -B_y & B_x & 0 \\
\end{array}
\right)$$ and the four-current $$J^\mu=(\rho,j_x,j_y,j_z)$$ The four-force on a charge distribution is then given by $f_\mu = J^\nu F_{\mu\nu} $. These equations hold in any reference frame.
Now, in your case you have a neutral wire with a steady current in the $x$ direction lying along the $x$ axis. This wire has a four-current of $$J^{\mu }=(0,i,0,0)$$  There is also a test charge of magnitude $q$ which is located a distance of $r$ away from the origin in the $z$ direction. This test charge has a four current of $$q^\mu=(q,0,0,0)$$
At the location of the test charge, the current produces the EM tensor $$F^{\mu \nu }=\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & -\frac{i}{r} \\
 0 & 0 & 0 & 0 \\
 0 & \frac{i}{r} & 0 & 0 \\
\end{array}
\right)$$ So the force on the test charge is $$f_{\mu }=q^{\nu }F_{\mu \nu } =(0,0,0,0)$$ which, as expected, means that there is no force on the charge in the rest frame.
Now, when we boost everything to a primed frame moving at $v$, represented by primes on the indices, we get the following: $$ J^{\mu '}=(\gamma  i v,\gamma  i,0,0) $$ where the non-zero first term indicates that the wire is no longer neutral in the primed frame, but has some net charge.  $$q^{\mu '}=\left(q',j',0,0\right)=\left(\gamma  q,\gamma  q v,0,0\right)$$ where the non-zero second term indicates that the test charge also has some "test current" because it is moving in this frame. And $$F^{\mu ' \nu '}=\left(
\begin{array}{cccc}
 0 & 0 & 0 & -\frac{i v \gamma }{r} \\
 0 & 0 & 0 & -\frac{i \gamma }{r} \\
 0 & 0 & 0 & 0 \\
 \frac{i v \gamma }{r} & \frac{i \gamma }{r} & 0 & 0 \\
\end{array}
\right)$$ where the new terms indicate the electric field produced by the charge of the wire in the primed frame at the location of the test charge. In this frame the force is calculated as $$f_{\mu '}=q^{\nu '} F_{\mu ' \nu '}=\left( 0,0,0,\frac{\gamma  i \left(j'-v
   q'\right)}{r}\right)=(0,0,0,0)$$ Notice that regardless of $v$ the term of force from the wire's electric field acting on the test charge, $\gamma i v q'/r$, is exactly canceled out by the term of force from the wire's magnetic field acting on the test current, $\gamma i j'/r$. So if the force is 0 in one frame it is 0 in all frames.
A: If you want to know why moving magnet doesnt affect moving charge, if their relative speed is zero:
You can think of it as if they both emit EM field. Moving magnet emitts magnetic and electric fields, mostly magnetic while speed is significantly below c. And moving charge emits electric and magnetic fields, mostly electric while speed is significantly below c.
As you take speed higher and higher as a frame of reference, but their relative speed remains zero, you can measure strong EM fields in your frame of reference, from either of them. But fields cancel out on their frame of reference, where their relative speed is zero.
