You don't need to know $m$. You know $m^3$ modulo each modulus, which is sufficient. You want to find:
$$c \equiv m^3 \pmod{n_b}$$
$$c \equiv m^3 \pmod{n_c}$$
$$c \equiv m^3 \pmod{n_d}$$
Because $n_b$, $n_c$, $n_d$ are pairwise coprime (assume they have no common factors) a solution must exist.
The Wikipedia page has a nice explanation of the algorithm to find $c$. The actual expression is:
$$c = c_b (n_c \cdot n_d) [ (n_c \cdot n_d)^{-1} ]_{n_b} + c_c (n_b \cdot n_d) [ (n_b \cdot n_d)^{-1} ]_{n_c} + c_d (n_b \cdot n_c) [ (n_b \cdot n_c)^{-1} ]_{n_d}$$
Where $[a^{-1}]_b$ is the multiplicative inverse of $a$ modulo $b$. Note $\gcd{(a, b)} = 1$ is always satisfied. Also, I used the notation $c_b = m^3 ~ \text{mod} ~ n_b$, $c_c = m^3 ~ \text{mod} ~ n_c$, $c_d = m^3 ~ \text{mod} ~ n_d$.
Let's try with some numbers. Suppose someone sends the message $m = 102$ to three different people with textbook RSA, with moduli $n_b = 377$, $n_c = 391$ and $n_d = 589$. So:
$$c_b = 102^3 ~ \text{mod} ~ 377 = 330$$
$$c_c = 102^3 ~ \text{mod} ~ 391 = 34$$
$$c_d = 102^3 ~ \text{mod} ~ 589 = 419$$
So the attacker wants to solve the following system of congruences:
$$c \equiv 330 \pmod{377}$$
$$c \equiv 34 \pmod{391}$$
$$c \equiv 419 \pmod{589}$$
Using the equation above, we obtain (compute each term separately for clarity):
$$t_b = c_b (n_c \cdot n_d) [ (n_c \cdot n_d)^{-1} ]_{n_b} = 330 (391 \times 589) [ (391 \times 589)^{-1}]_{377} = 24471571740$$
$$t_c = c_c (n_b \cdot n_d) [ (n_b \cdot n_d)^{-1} ]_{n_c} = 34 (377 \times 589) [ (377 \times 589)^{-1}]_{391} = 505836734$$
$$t_d = c_d (n_b \cdot n_c) [ (n_b \cdot n_c)^{-1} ]_{n_d} = 419 (377 \times 391) [ (377 \times 391)^{-1}]_{589} = 35452267942$$
$$\therefore c = t_b + t_c + t_d ~ \text{mod} ~ (n_b \cdot n_c \cdot n_d) = 1061208$$
And we get $m = \sqrt[3]{c} = \sqrt[3]{1061208} = 102 = m$.
Note this attack on textbook RSA would work on any exponent given a sufficiently large number of people, of course $e = 3$ is the most realistic setting (and easiest to demonstrate, for obvious reasons).
The idea, of course, being to use these relations with the CRT to manufacture a relation of the form $m^3 \equiv x \pmod{n'}$ where $n'$ is on the order of $n^3$ (and, more crucially, where $1 < m < n$) such that $m^3$ is not reduced (and so you can just take cube roots).