Your method of solution is correct. Your mistakes (I think) are in the calculations of $T_1, T_2$ and $x$, all of which should depend on angle $\theta$. Your formula for $T_2$ depends on $\theta$ but that for $T_1$ does not. Likewise your formula for $x$ appears to be $2R_1-2R_2\sin\theta$. The 2nd term depends on $\theta$ but the 1st term does not.

The motion in each region is a circular arc of radius $r$. The centripetal force for this motion is provided by the magnetic force on the particle : $$\frac{mv^2}{r}=Bqv$$ $$r=\frac{mv}{qB}$$ The distance moved along the $x$ axis is the chord of the arc, which is $2r\sin\theta$ where $\theta$ is the angle between the arc and the $x$ axis.

After tracing an arc in both regions, the motion repeats. During each cycle the resultant distance moved in the $+x$ direction is $$x=2(r_1-r_2)\sin\theta=\frac{2mv\sin\theta}{q}(\frac{1}{B_1}-\frac{1}{B_2})=\frac{2mv\sin\theta (B_2-B_1)}{qB_1B_2}$$

The angle subtended by each arc is $\theta_1=2\theta$ and $\theta_2=2(\pi-\theta)$ respectively. The times for the particle to traverse each arc are $\frac{r_1\theta_1}{v}, \frac{r_2\theta_2}{v}$ respectively. The total time for each cycle of the motion is therefore $$t=\frac{1}{v}(r_1\theta_1+r_2\theta_2)=\frac{1}{v}(2\theta\frac{mv}{qB_1}+2(\pi-\theta)\frac{mv}{qB_2})=\frac{2m((\pi-\theta) B_1+\theta B_2)}{qB_1B_2}$$

The average velocity of the particle in the $+x$ direction is $$V=\frac{x}{t}=\frac{(B_2-B_1)v\sin\theta}{(\pi-\theta)B_1+\theta B_2}=\frac{(B_2-B_1)v\sin\theta}{\pi B_1+\theta (B_2-B_1)}$$

edited Aug 9, 2018 by sammy gerbil