I will use variances instead of standard deviations in notation for random variables.
For any two random variables $A$ and $B$ with respective means $\mu_A, \mu_B$ and respective variances $\sigma_A^2, \sigma_B^2$, what is the "best" estimate of $B$ in terms of $A$? By "best" we mean that if $g(A)$ denotes the estimate of $B$ in terms of $A$, then we seek the function $g(\cdot)$ such that $E[(B-g(A))^2]$, the mean-square error of the estimate, is as small as possible. It is well-known that the minimum mean-square error (MMSE) estimator of $B$ in terms of $A$ is the conditional mean $E[B\mid A]$ of $B$ given $A$. This conditional mean is a random variable and it is a function of $A$, not of $B$ as it seems, that is, $E[B\mid A]$ is the $g(A)$ that we seek. It is also known that $E[g(A)] = E\big[E[B\mid A]\big]$, the mean of this particular function of $A$, happens, by a miracle of modern mathematics, to equal $E[B] = \mu_B$, the mean of $B$.
A more constrained version of the above function optimization problem asks, "What is the linear minimum mean-square error (LMMSE) estimator of $B$ in terms of $A$" where we seek estimators that are constrained to be of the form $\alpha A + \beta$ and we try to find $\alpha$ and $\beta$ such that $E[(B-(\alpha A + \beta))^2]$ is as small as possible. Here the well-known answer is that
$$\alpha = \rho\frac{\sigma_B}{\sigma_a}, \quad \beta = \mu_B - \rho\frac{\sigma_B}{\sigma_a}\mu_A = \mu_B - \alpha\mu_A \tag{1}$$t
where $\rho$ is the correlation coefficient,
and it is also known that this function $\alpha A + \beta$ has mean and variance given by
$$E[\alpha A + \beta] = \alpha \mu_A + \beta = \mu_B, \quad
\operatorname{var}(\alpha A + \beta) = \sigma_B^2(1-\rho)^2.\tag{2}$$
As pointed out by @Ben, your assumptions that $A$ is a normal random variable $N(0, \sigma_A^2)$ and that the conditional distribution of $B$ given that $A$ has value $a$ is $N(qa,\sigma_b^2)$ (where $q$ and $\sigma_b$ are known constants (not dependent on $a$ in any way)) is equivalent to the assumption that $A$ and $B$ are jointly normal random variables. If you don't like this bald assertion, notice that
\begin{align}
f_A(a) &\propto \exp\left(-\frac{a^2}{2\sigma_A^2}\right)\\
f_{B\mid A}(b\mid a) &\propto \exp\left(-\frac{(b-qa)^2}{2\sigma_b^2}\right)
\end{align}
and so the joint density $$f_{A,B}(a,b) = f_{B\mid A}(b\mid a)\cdot f_A(a) \propto \exp(Q(a,b))$$
where $Q(a,b)$ is a quadratic function of $a$ and $b$, that is, after completing the square etc, it will be found that the joint density is a jointly normal (a.k.a. bivariate normal) density.
Now, in the special case when $A$ and $B$ are jointly normal random variables (that is, they have a bivariate normal density), the MMSE estimator $E[B\mid A]$ is a linear function of $A$, and so the MMSE estimator must coincide with the LMMSE estimator, no? For jointly normal random variables, the
conditional distribution of $B$ given $A = a$ is a normal distribution with mean and variance given by
$$E[B \mid A = a] = \rho\frac{\sigma_B}{\sigma_A}a + \mu_B - \rho\frac{\sigma_B}{\sigma_a}\mu_A, \quad
\operatorname{var}(B \mid A = a) = \sigma_B^2(1-\rho^2) \tag{3}$$
where we are told that $\mu_A = 0$ in this instance, and that $$E[B \mid A = a] = qa, \quad \operatorname{var}(B \mid A = a) = \sigma_b^2\tag{4}$$ (Note the subscript is a lower-case $b$). Comparing $(3)$ and $(4)$ we see that it must be that $\mu_B = 0$, and that the known $q$ equals $\rho\frac{\sigma_B}{\sigma_A}$ (remember that we are given the value of $\sigma_A$) while the known $\sigma_b^2$ must equal $\sigma_B^2(1-\rho^2)$.
So we get that
$$
\rho\sigma_B = q\sigma_A, \sigma_b^2 = \sigma_B^2-\rho^2\sigma_B^2
= \sigma_B^2 - (q\sigma_A)^2\\
\implies
\sigma_B^2 = \sigma_b^2 + (q\sigma_A)^2, \quad \rho = \frac{q\sigma_A}{\sqrt{\sigma_b^2 + (q\sigma_A)^2}} \tag{5}
$$
So, $A$ and $B$ are zero-mean jointly normal random variables with variances $\sigma_A^2$ and $\sigma_B^2 = \sigma_b^2 + (q\sigma_A)^2$ and correlation coefficient $\rho = \frac{q\sigma_A}{\sqrt{\sigma_b^2 + (q\sigma_A)^2}}$. It follows that the conditional distribution of
$A$ given that $B = b$ is a normal distribution with mean and variance
$$E[A \mid B = b] = \rho\frac{\sigma_A}{\sigma_B}b, \quad
\operatorname{var}(A \mid B = b) = \sigma_A^2(1-\rho^2)$$
