Any number $x \in \mathbb{Z}$ can be written as $\pm\sum\limits_{i=0}^n x_i2^i$ where $x_i \in \text{{0,1}}$ (signed binary representation). It can also be written as $\sum\limits_{i=0}^n x_i(-2)^i$ where $x_i \in \text{{0,1}}$ (negabinary representation).
We are given two functions $B(x)$ and $N(x)$, where $B(x)$ interprets the binary representation of the decimal input as a negabinary input, which it converts to binary and then to decimal and where $N(x)$ converts the binary representation of the decimal input to negabinary, which is then interpreted as binary and then as decimal.
For example:
$B(3_{10}) \rightarrow B(11_{-2}) = -1_{2}=-1_{10}$
and
$N(2_{10}) = N(10_{2})=110_{-2} \rightarrow 110_2=6_{10}$
Now the following is true:
$\sum\limits_{i=1}^{\infty} \frac{1}{B(2n+1)^2}=\frac{\pi^2}{4}$
because $B(2n+1)^2$ contains the square of all odd numbers twice and the sum of the recipricals of all odd numbers is $\sum\limits_{i=1}^{\infty} \frac{1}{(2n+1)^2}= \frac{\pi^2}{8}$
The question now is, what is $\sum\limits_{n=1}^{\infty} \frac{1}{N(n)^2}$ equal to?
As you've noticed, B is a bijection between positive odd numbers and all odd numbers.
N's image is the set of integers whose binary string when interpreted in negabinary are positive. A binary string will be positive iff it's length is odd, since the leading term dominates the sign of the string (since $2^n>2^{n-1}+2^{n-2}...$). Thus, you want $1/1^2 + 1/4^2+1/5^2 + 1/6^2 + 1/7^2 + 1/16^2 + 1/17^2 ... + 1/31^2 + 1/64^2 ...$.
I don't know of a closed form for this series, but it's been discussed separately here: Infinite sum of odd-length binary numbers