This is asking for the value of an $L$-function of an even Dirichlet character $\chi$ at a positive even integer, and these have known values. It is analogous to the explicit expressions for the Riemann zeta-function at positive even integers (definitely not at positive odd integers!), but instead of the values being rational multiples of powers of $\pi$, they are *algebraic* multiples of powers of $\pi$.

The particular Dirichlet character $\chi$ in this question is defined modulo 10, but it is a lifting of a character mod 5. Let $\psi(n) = (\frac{n}{5})$, which is a nontrivial even Dirichlet character mod 5. For odd $n$, $\chi(n) = \psi(n)$, so for ${\rm Re}(s) > 1$,
$$
L(s,\chi) = \sum_{n \geq 1} \frac{\chi(n)}{n^s} = \prod_{p > 2} \frac{1}{1 - \psi(p)/p^s} = \left(1 - \frac{\psi(2)}{2^s}\right)L(s,\psi) =
\left(1 + \frac{1}{2^s}\right)L(s,\psi).
$$
Taking $s = 2$,
$$
\sum_{n \geq 1} \frac{\chi(n)}{n^2} = \frac{5}{4}L(2,\psi).
$$
I will show $L(2,\psi) = \frac{4}{25\sqrt{5}}\pi^2$, and multiplying this by $5/4$ gives the desired value for $\sum \chi(n)/n^2$.

The character $\psi \bmod 5$ is primitive since every nontrivial Dirichlet character modulo a prime is primitive. It is a quadratic character: its nonzero values are $\pm 1$.
For each *even* primitive quadratic Dirichlet character $\eta \bmod m$, let's work out a formula for $L(k,\eta)$ when $k$ is a positive *even* integer and then apply it to $\eta = \psi$ and $k = 2$. (There are analogous formulas for $L(k,\eta)$ when $\eta \bmod m$ is an *odd* primitive quadratic Dirichlet character and $k$ is a positive *odd* integer, and also formulas when $\eta$ is not quadratic, but I omit all of this for simplicity.)

For even primitive quadratic $\eta \bmod m$ and ${\rm Re}(s) > 1$, the completed $L$-function of $\eta$ is
$$
\Lambda(s,\eta) := \left(\frac{\pi}{m}\right)^{-s/2}\Gamma\left(\frac{s}{2}\right)L(s,\eta).
$$
This turns out to be an entire function, so $L(s,\eta)$ is also entire, and we have a functional equation $\Lambda(s,\eta) = \Lambda(1-s,\eta)$. (If $\eta$ were not quadratic, the functional equation would be more complicated.) When $s = k$ is a positive even integer, unraveling the formula $\Lambda(k,\eta) = \Lambda(1-k,\eta)$ gives us
$$
\left(\frac{\pi}{m}\right)^{-k/2}\Gamma\left(\frac{k}{2}\right)L(k,\eta) =
\left(\frac{\pi}{m}\right)^{-(1-k)/2}\Gamma\left(\frac{1-k}{2}\right)L(1-k,\eta),
$$
so
$$
L(k,\eta) = \left(\frac{\pi}{m}\right)^{k - 1/2}\frac{\Gamma((1-k)/2)}{\Gamma(k/2)}L(1-k,\eta).
$$
The first factor on the right is a known expression. Using the reflection and duplication formulas for the Gamma-function,
$$
\frac{\Gamma((1-s)/2))}{\Gamma(s/2)} = \frac{2^{s-1}\sqrt{\pi}}{\cos\left(\frac{\pi}{2}s\right)\Gamma(s)}.
$$
Letting $s = k$ be an even integer, $\cos((\pi/2)k) = \cos((k/2)\pi) = (-1)^{k/2}$, so
$$
L(k,\eta) = \left(\frac{\pi}{m}\right)^{k - 1/2}\frac{(-1)^{k/2}2^{k-1}\sqrt{\pi}}{(k-1)!}L(1-k,\eta) =
\frac{(-1)^{k/2}2^{k-1}}{(k-1)!m^{k-1/2}}L(1-k,\eta)\pi^k
$$
and the number $L(1-k,\eta)$ for positive integers $k$ and nontrivial primitive $\eta \bmod m$ turns out to be *algebraic*. (That generalizes the Riemann zeta-function at negative integers being *rational*). Explicitly,
$$
L(1-k,\eta) = -\frac{B_{k,\eta}}{k}
$$
where the "twisted" Bernoulli number $B_{k,\eta}$ has exponential generating function
$$
\sum_{k \geq 0} B_{k,\eta}\frac{x^k}{k!} =
\sum_{j=1}^m \eta(j)\frac{xe^{jx}}{e^{mx}-1},
$$
so $B_{k,\eta}$ is a $\mathbf Q$-linear combination of values of $\eta$.
(See Larry Washington's "Introduction to Cyclotomic Fields", esp. Theorem 4.2.)

Let's specialize all this to the case when $\eta$ is the (even primitive) quadratic character $\psi \bmod 5$ above and $k = 2$. We get
$$
L(2,\psi) = \frac{-2}{5^{3/2}}L(-1,\eta)\pi^2 =
\frac{B_{2,\psi}}{5\sqrt{5}}\pi^2.
$$
We had said the coefficient of $\pi^2$ turns out to be $4/(25\sqrt{5})$, and we'll get this by showing $B_{2,\psi} = 4/5$.

For the character $\psi \bmod 5$,
$$
\sum_{k \geq 0} \frac{B_{k,\psi}}{k!}x^k = \frac{x(e^x - e^{2x}-e^{3x}+e^{4x})}{e^{5x}-1} = \frac{xe^x(e^x-1)^2(e^x+1)}{e^{5x}-1} =
\frac{2}{5}x^2 - \frac{1}{3}x^4 + \cdots,
$$
so looking at the coefficient of $x^2$ on both sides tells us
$B_{2,\psi} = 4/5$.