With [tex]2L=\pi[/tex], the Fourier series expansion of [tex]f(x)[/tex] is
[tex]\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L[/tex]
[tex]\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx[/tex]
where the coefficients are obtained by computing
[tex]\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx[/tex]
[tex]\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx[/tex]
[tex]\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx[/tex]
[tex]\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx[/tex]
[tex]\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx[/tex]
[tex]\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx[/tex]
You should end up with
[tex]a_0=0[/tex]
[tex]a_n=0[/tex]
(both due to the fact that [tex]f(x)[/tex] is odd)
[tex]b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)[/tex]
Now the problem is that this expansion does not match the given one. As a matter of fact, since [tex]f(x)[/tex] is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.
One possibility is that you're actually supposed to use the even extension of [tex]f(x)[/tex], which is to say we're actually considering the function
[tex]\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3<|x|\le\frac{2\pi}3\\-\frac\pi3&\text{for }\frac{2\pi}3<|x|\le\pi\end{cases}[/tex]
and enforcing a period of [tex]2L=2\pi[/tex]. Now, you should find that
[tex]\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)[/tex]
The value of the sum can then be verified by choosing [tex]x=0[/tex], which gives
[tex]\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)[/tex]
[tex]\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots[/tex]
as required.
