Block diagram reduction
I believe that it will be worth for many people to do it from scratch. Let's call the signal after input and feedback summation $E(s)$:



Then question arises what $E(s)$ is equal to? Well that's simple:

$$E(s)=5R(s)-(5+K_vs)C(s) $$

So we have our backward path, now the forward one. What $C(s)$ is equal to?

$$C(s)=E(s)K_p \dfrac{2}{(Js+B)s} $$

Let's substitute previously calculated $E(s)$ to the above equation:

$$C(s)=\dfrac{2}{(Js+B)s}K_p[5R(s)-(5+K_vs)C(s)] $$

Because we have our output $C(s)$ on both sides of the equation, we must eliminate that. Expand the right side of the equation:

$$C(s)=\dfrac{2}{(Js+B)s}5K_p\cdot R(s)-\dfrac{2}{(Js+B)s}K_p(5+K_vs)C(s) $$

Now move the part with $C(s)$ to the left side and move it before parenthesis, this yields:

$$C(s) \left[1+ \dfrac{2}{(Js+B)s}K_p(5+K_vs)\right]=\dfrac{2}{(Js+B)s}5K_p\cdot R(s) $$

Now you can divide right side of the equation by the factor from the left side:

$$C(s) =  \dfrac{\dfrac{2}{(Js+B)s}5K_p}{1+ \dfrac{2}{(Js+B)s}K_p(5+K_vs)}\cdot R(s) $$

Great! Now we have direct relation between the input and the output. Let's make this fraction more pleasant:

$$C(s) =  \dfrac{\dfrac{10K_p}{(Js+B)s}}{1+ \dfrac{10K_p+2K_pK_vs}{(Js+B)s}}\cdot R(s) $$

Now expand the one in denominator to: $\dfrac{(Js+B)s}{(Js+B)s} $, and add it:

$$C(s) =  \dfrac{\dfrac{10K_p}{(Js+B)s}}{\dfrac{(Js+B)s + 10K_p+2K_pK_vs}{(Js+B)s}}\cdot R(s) $$

We can now remove common $(Js+B)s$:

$$C(s) =  \dfrac{10K_p}{(Js+B)s + 10K_p+2K_pK_vs}\cdot R(s) $$

Finally let's do one last multiplication:

$$C(s) =  \dfrac{10K_p}{Js^2+Bs + 10K_p+2K_pK_vs}\cdot R(s) $$

And voila!

$$C(s) =  \dfrac{10K_p}{Js^2+ s(B +2K_pK_v)+ 10K_p}\cdot R(s) $$